Section 1:   Linear transformations and arrays

Functions may act on arrays. For example,

Reccall that arrays can be represented as vectors. If u = (x,y)   and   v = (α,β) then

Let v represent an array. Then

[1.1] (Linear transformation) A linear transformation L on a vector space of arrays all of the same length is a function satisfying the following two conditions:
  (a) L(u + v) = L(u) + L(v)
  (b) L(λv) = λL(v)
where u and v are vectors (arrays) and λ is any real number.

In most discussions, the lengths of arrays will be 2 or 3, but in most discussions the facts will apply to longer arrays as well. The term vector space will mean the collection of all arrays of length 2, or the collection of all arrays of length 3.
Not discussed are linear transformations between vector spaces of arrays of different lengths, for example, linear transformations from the collection of arrays of length 3 to the collection of arrays of length 2.
The abbreviation lintr will mean "linear transformation". Also the letter L will be used exclusively for lintr. If there are more than one lintr in a discussion, then subscripts will be used: L₁, L₂, L₃,... .

Recall that the zero vector 0 is an array of zero coordinates:   0 = (0,0)   or   0 = (0,0,0).
The following states a necessary condition that a function be a lintr:

[1.2] (All Linear transformations leave the zero vector fixed) Every linear transformation carries the zero vector back onto itself.
Notation: L(0) = 0

Part (b) of [1.1] provides a simple proof. Set λ = 0.

What makes any lintr useful is its effect on lincombs (linear combinations) of vectors:

[1.3] (Linear transformations of linear combinations) A linear transformation carries a linear combination of vectors onto a linear combination of images of those vectors but with the same coefficients.
Notation: L(αa + βb + γc + δd + ...) = αL(a) + βL(b) + γL(c) + δL(d) + ...

This statement is actually an application of definition [1.1] many times to the terms of the lincomb.

Recall that the orthonormal arrays (1,0), (0,1) and (1,0,0), (0,1,0), (0,0,1) are special arrays, because for any arbitrary arrays (x,y) and (x,y,z) the letters may be "factored out" of the arrays leaving the special arrays with the letters as coefficients:

If images L(1,0) and L(0,1) are known, then the image L(x,y) can be computed. For example, if

[1.4a] (Images of special arrays) A linear transformation is completely determined by the images of the special arrays.
Notation: if images L(1,0) and L(0,1) are known, then L(x,y) can be computed from xL(1,0) + yL(0,1)
If images L(1,0,0), L(0,1,0) and L(0,0,1) are known,
     then L(x,y,z) can be computed from xL(1,0,0) + yL(0,1,0) + zL(0,0,1).

[1.4b] (Images of special arrays)
If   L(1,0) = (α₁, α₂)   and   L(0,1) = (β₁, β₂)   then

Click here to see the computations supporting these equalities.
The reader can verify that if

Expressions like 2x + 4y - 2z,   x + 5y + 7z,   4x + 6y + 8z occur often as coordinates.
An expression αx + βy is called a linear expression in x and y. An expression αx + βy + γz is called a linear expression in x,y and z. Notice that a linear expression has no vectors, just real numbers and variables x,y and possibly z. And every term has x,y or z in it. Linear expressions are used to determine linear transformations expressed in terms of coordinates.

[1.5] (Linear transformations and linear expressions)
If L(x,y) = (linear expression in x and y, linear expression in x and y) then L is a linear transformation on arrays of length 2.
If L(x,y,z) = (linear expression in x,y and z, linear expression in x,y and z, linear expression in x,y and z) then L is a linear transformation on arrays of length 3.

For example,

Through the use of images of special arrays, there is a close association between linear transformations and matrices:

[1.6] (Associated matrices) For any linear transformation, its images of the special arrays become columns in the associated matrix.
Notation:
   (a) if   L(1,0) = (α₁, α₂)   and   L(0,1) = (β₁, β₂)   then the following matrix M is associated with L:

The matrices contain all of the coordinates of the images of the special arrays. However, the when linear transformations are involved, the arrays are written horizontally and separately: for 2x2 arrays, (α₁, α₂)   and   (β₁, β₂); for 3x3 arrays (α₁, α₂, α₃),   (β₁, β₂, β₃)   and   (γ₁, γ₂, γ₃). But in the matrices the images are written in columns: for 2x2 arrays: (α₁, α₂)^t   and   (β₁, β₂)^t; for 3x3 arrays, (α₁, α₂, α₃)^t,   (β₁, β₂, β₃)^t   and   (γ₁, γ₂, γ₃)^t, where ( )^t is the transpose of ( ).

Iintuitively speaking, the row of αs become the (first) column of αs, the row of βs become the (second) column of βs, and if present the row of γs becomes the (third) column of γs.

GIven a linear transformation defined by L(x,y) or L(x,y,z), the associated matrix is easily found. For example, what is the matrix associated with the linear transformation L defined by L(x,y) = (5x + 6y, 8x + 7y) ?
First needed are the images of the special arrays. L(1,0) = (5,8) [set x=1, y=0],   L(0,1) = (6,7) [set x=0, y=1]. The columns of the matrix M associated with L are (5,8)^t and (6,7)^t. Therefore the matrix is:

Given the matrix M, is it possible to produce the lintr L, or more exactly, produce L(x,y) ? The answer is yes. But it must be remembered that with M vertical arrays are used, while with L, horizontal arrays are used.

[1.7] Let M be the matrix associated with a linear transformation L.
   (a) For the vector space of 2x2 arrays,   M(x,y)^t = L(x,y)^t
   (b) For the vector space of 3x3 arrays,   M(x,y,z)^t = L(x,y,z)^t

From [1.7] it may be said that the matrix of a lintr has enough of the essential information of the lintr. Since the associated matrix of a lintr on a vector space is square (2x2 or 3x3) it is possible to give meaning to the phrase the determinant of a linear transformation. The determinant of a linear transformation on a vector space is the determinant of the associated matrix.

Recall that it is sometimes possible to combine two functions called their product. If L₁ and L₂ are linear transformations, then it is possible to form their product L₂L₁. If L₁ carries u onto v and L₂ carries v onto w then L₂L₁ carries u onto w.
Notation: L₂L₁(u) = L₂(v) = w

Notice that the product is written in the reverse order of their execution: in L₂L₁ the lintr L₁ acts before L₂ acts.

[1.8] (Products of linear transformations) The product of two linear transformations (on the same vector space) is again a linear transformation (on that vector space).
Notation: L₁ and L₂ are linear transformations, then L₂L₁ is a linear transformation.

Click here to see an argument supporting this statement.

[1.9] (Products of linear transformations and their associated matrices) The matrix associated with the product (composition) of two linear transformations (on the same vector space) is equal to the product of the matrices associated with those linear transformations, taken in the same order.
Notation: If M₁ and M₂ are the matrices associated with the linear transformations L₁ and L₂, then the product M₂M₁ is the matrix associated with the product L₂L₁.

Click here to see proofs for arrays of lengths 2 and 3.

This theorem [[1.9] can be used to find a simple way for computing the product of two linear transformations: shift the work to computing the product of their associated matrices.
Suppose L₁ and L₂ are linear transformations on 2x2 arrays defined by

The identity functions I₂ and I₃ on arrays of lengths 2 and 3 respectively carries each array back onto itself (leaves every array fixed):

For any lintr L,   LI = L   and   IL = L.   Similar statements may be said about the matrix M associated with L.
The introduction of the identity function allows the question   does a lintr L have an inverse? Is there a linear transformation L^-1 such that LL^-1 = I? The following answers this question.

[1.10] (Inverse of a linear transformation) Let L be a linear transformation on a vector space. If   det L   is not zero, then an inverse linear transformation   L^-1   exists. The matrix associated with   L^-1   is the inverse of the matrix associated with L.

Recall that by definition   det L = det M   where M is the matrix associated with L. Therefore the hypothesis   det L is not zero   means that M has an inverse. From the inverse of M a linear transformation L* can be constructed. Since MM^-1 = I the product   LL* = I.  . This makes L* the inverse of L, that is L* = L^-1.

[1.11] A linear transformation on a vector space of arrays is called singular if its determinant is zero. A linear transformation on a vector space of arrays is non-singular if its determinant is not zero. It is not difficult to show that all non-singular linear transformations on the same vector space of arrays form a group called the full linear group.

Section 2:   Linear transformations and geometric points, lines and the plane

Let L be a lintr. Quite often the symbol A', B', C', P', Q', R' are symbols used to denote is the image of points A, B. C, P, Q, R:

Example: Suppose L is defined by L(x,y) = (2x + 3y, 4x + 5y). If P(6,5) then P'(27,49) because L carries array (6,5) onto array (27,49). For this example,   x' = 2x + 3y   and   y' = 4x + 5y.

When the term "image" is used it means that some lintr is involved to produce that image.
The following statement reveals a simple but important property of linear transformations.

[2.1] (The origin left fixed) Every linear transformation must carry the origin back onto itself.
Notation: L(O) = O.

This is a direct result of the fact that a lintr carries the array 0 of all zeros back onto itself. Depending upon the lintr there may or may not be other points carried onto the origin.

If P' is the image of point P then the images of O and P are O and P', that is, L(OP) = OP'. This says that position vector p is carried onto position vector p':   L(p) = p'.

[2.2] (Images of position vectors) A linear transformation carries a position vector locating a point onto a position vector locating the image of that point.
Notation: if position vector p locates point P then L(p) is a position vector that locates the point L(P).
L(p) = p' where p' locates P'.

Recall that collinear points lie on a common line. The use of position vectors provides simple argument for the following fact:

[2.3] (Linear transformations preserve collinearity) Every linear transformation carries collinear points onto collinear points.
Notation1: if A,B,P are collinear points then L(A), L(B), L(P) are collinear points.
Notation2: if A and B are distinct, and if P is a point on the line through A and B, then image L(P) is on the line through L(A) and L(B).

IF A and B coincide, then L(A) and L(B) coincide and [2.3] is trivial. Suppose A and B are distinct. Using Notation2, AP = λAB for some real number λ. This leads to the equation involving position vectors:

By [2.3] all linear transformations carry collinear points onto collinear points. But what happens if the points are not collinear? If points A,B,C are not collinear, what about their images L(A), L(B), L(C) ?

[2.4] (Images of points which are not collinear) A non-singular linear transformation carries non-collinear points onto non-collinear points.
Notation: if points A,B,C are not collinear and L is non-singular, then images L(A),L(B),L(C) are not collinear.

Click here to see the proof of this statement.

[2.5] (images of segment) A linear transformation carries an entire segment onto a segment or onto a point.
Notation: L carries all the points on a segment AB onto all the points on a segment with end points L(A) and L(B).

Click here to see the proof of this statement. Statements [2.4] and [2.5] imply that a non-singular lintr carries a genuine triangle onto a genuine triangle, more generally a polygon onto a polygon. This gives lintr a geometric character.
The proof of [2.3] can be adapted to prove the following:

[2.6] (Image of line) A linear transformation carries an entire line onto an entire line or onto a point..

Click here to see discussion supporting this statement.
Lines are (straight) subsets of the plane. The following statements list the possible images of the entire plane.

[2.7] (Images of the plane) [sp3] A linear transformation on the plane does one of the following:
   (a) [rank 2] it carries the entire plane onto itself (the entire plane is also the image);
   (b) [rank 1] it carries the entire plane onto a line through the origin (the entire line is the image and is in the plane);
   (c) [rank 0] it carries the entire plane onto the origin.

The rank number assigned to the lintr is actually the dimension of image of the plane. Planes, lines, points have dimensions 2, 1, 0 respectively. Also they are the the ranks of the 2x2 matrices associated with the transformation.
For lintr of rank 0 see [2.11] below.    For lintr of rank 1 see [2.13] below.    For lintr of rank 2 see [2.14] below.

There is only one lintr of rank 0 ([2.7c]). It is considered trivial and not interesting. Squashing the entire plane into a single point removes anything interesting. Discussions below involve mostly non-trivial lintrs. On the plane, only lintrs of rank 2 ([2.7a]) are non-singular. Intuitively speaking, only they can preserve recognizable shapes of geometric figures such as parallelograms and triangles (however they may be resized and/or sheared). A lintr of rank 1 will send any geometric figure into a line (through the origin) and flaten it inside the line. Transformations of rank 1 and rank 0 are singular, and never have inverses. Intuitively speaking, their effects on the plane cannot be undone by some reverse transformation.

Let L be any lintr on the plane. Then L can do one of three things with a point P:
  (a) L carries P onto the origin O:   L(P) = O.
  (b) L carries P back onto itself:   L(P) = P.
  (c) L carries P onto some point in the plane that is not O nor P.
In an intuitive sense, cases (a) and (b) are special, and therefore give P a special status with L.

Part (a) leads to the idea of kernels discussed in section 6.
Part (b) leads to the idea of fixed points discussed in section 5.

Suppose A, B, C, D are four points satisfying the vector equation

[2.8] (Preserving proportionality) If one vector is λ times a second vector, then for any linear transformation, the image of the first vector is λ times the image of the second vector, where λ is any real number.
Notation: If vector AB = λCD then vector A'B' = λC'D'.

Note: [2.8] does not imply that vectors AB and its image A'B' are equal, have the same length or are even being parallel. The lintr may carry a horizontal or vertical figures onto images at any slant or angle.
From [2.8] it is possible to deduce the following, which is in the more traditional geometric language:

[2.9] (Preserving parallelism and proportionality) For any linear transformation,
  (a) it carries parallel segments onto parallel segments;
  (b) it carries parallel lines onto parallel lines;
  (c) the ratio of the lengths of the segments is equal to the ratio of the lengths of the images of those segments.
  (d) the orientation of the segments is equal to the orientation of their images (if vectors AB and CD point in the same direction [or opposite directions], then vectors A'B' and C'D' point in the same direction [or opposite directions] respectively).

To simplify the wording of [2.9b] allow a line to be parallel to itself. This allows singular lintrs to have the property described in [2.9b]. This convention also makes the relation "is parallel to" an equivalence relation between lines.
The following is a noteworthiy link between linear algebra and geometry.

[2.10] (preserving parallelograms and trapezoids) For any non-singular linear transformation,
  (a) it carries a parallelogram onto a parallelogram;
  (b) it carries a trapezoid onto a trapezoid.

Click here to see more discussion of lintrs preserving parallelograms and trapezoids. However, there exist some lintrs that do not preserve squares, rectangles or isosceles trapezoids.


The special arrays (1,0 and (0,1) become here special points. Any point (x,y) is a linear combination of these points:

For example, suppose a linear transformation L carries the special points onto the origin.

This proves the following:

[2.11] (Rank 0) A linear transformation carries every point in the plane onto the origin if and only if it carries the special points onto the origin.
Notation1: L(P) = O for every point P in the plane
Notation2: L(x,y) = (0,0) for all real values of x and y.

The lintr in [2.11] is trivial. It is often called the

Similarly, using L(1,0) = (1,0)   and   L(0,1) = (0,1) it is easy to show L(x,y) = (x,y). Therefore,

[2.12] (Leaving points fixed) A linear transformation leaves every point fixed (it is the identity transformation) if and only if it leaves the special points fixed.

The following says that if a lintr carries the special points onto a line (the line must pass through the origin because O is always an image) then the lintr carries the entire plane onto that line.

[2.13] (Rank 1) A linear transformation is not the zero transformation and carries every point onto a line through the origin   if and only if   the images of the special points are not both zero and both lie on a line through the origin.

Click here for a short proof of this statement.

Using the identity L(x,y) = xL(1,0) + yL(0,1) the reader can easily show that if the lintr carries the special point (1,0) onto the origin then the lintr carries the plane onto the y-axis: L(x,y) = (0,y).   Similarly, if it carries the point (0,1) onto the origin then the plane is carried onto the x-axis: L(x,y) = (x,0), Not both images are the origin.

[2.14] (Rank2) A linear transformation carries the entire plane back onto itself if and only if the images of the special points are not both on the same line (through the origin)..

Click here for a proof of this statement.

The linear transformation in [2.14] is non-singular.

The special triangle is a genuine triangle - its vertices are non-collinear points. Is its image a genuine triangle?

Sometimes statements are made better by using the special triangle. Let E₁ and E₂ denote the special points:   E₁(1,0),   E₂(0,1). These points and the origin are vertices of a special triangle OE₁E₂.

The special triangle has four important properties:
   (a) it has a vertex at the origin O;
   (b) the two sides OE₁ and OE₂ have lengths equal to 1 (so the triangle is isosceles);
   (c) it is a right triangle because two sides lie along the perpendicular x-axis and the y-axis:
     (angle E₁OE₂ = 90°);
   (d) the angle E₁OE₂ is measured in a positive (counter-clockwise direction): OE₁ is the initial side of the angle and OE₂ is the terminal side. Notice that the label OE₁E₂ gives the triangle a positive (counterclockwise) orientation.

Oriientations will involve only linear transformations of rank 2 (non-singular).
For a discussion on orientation click here.

Lying along the congruent sides of the special triangle are special vectors:

Intuitively speaking, what a linear transformation does to the special points determines what it does to the special triangle, and conversely. In particular, some linear transformations carry geometric figures onto congruent figures. The most simple siguation involves segments. A linear transformation that carries all segments onto congruent segments is said to preserve distance. The following is a more direct definition:

[2.15] (Preserving distance) A linear transformation L preserves distance if and only if the distance between any pair of points is equal to the distance between their images.
Notation1: Linear transformation L preserves distance   if and only if   distance between images L(P) and L(Q) is the same as the distance between any points P and Q.
Notation2: For any pair of points (x₁, y₁) and (x₂, y₂),    |L(x₁, y₁) -- L(x₂, y₂)| = |(x₁, y₁) -- (x₂, y₂)|.

Although the name is not generally used, a distance preserving transformation might be called a "congruence linear transformation." However, in this chapter on linear transformations, the phrases a "linear transformation that preserves distance" or "distance preserving linear transformation"will be used. (For readers who have heard the term "isometry", it will be introduced in the next chapter on affine transformations.)

Example 1
The identity transformation I preserves distance because it leaves everything fixed and unchanged. More specifically, I(P) = P and I(Q) = Q, so the distance between I(P) and I(Q) is the same as the distance between P and Q trivially.

Using absolute value notation and coordinates, L,, defined by L(x,y) = some formula in x and y, preserves distance if and only if

Example 2
The linear transformation L that interchanages coordinates   L(x,y) = (y,x)   preserves distance:
( distance between L(x₁,y₁) and L(x₂,y₂) ) squared    =
    |L(x₁,y₁) -- L(x₂,y₂)|²    =    |(y₁, x₁) -- (y₂, x₂)|²    =    ... [chain of equalities]...    =    |(x₁, y₁) -- (x₂, y₂)|² =
(distance beetween (x₁,y₁) and (x₂,y₂) squared .
Click here to see the complete chain of equalities.

The proof of [2.15] and Example 2 could be shortened by using the following:

[2.16] (Preserving norms) A linear transformation preserves distance if and only if it carries every position vector onto a position vector of equal length.
Notation 1: L preserves distance if and only if |L(p)| = |p| for every position vector p.
Notation 2: L preserves distance if and only if |L(x,y)| = |(x,y)| for every point (x,y).

Assume that L preserves distance. To prove that it carries position vectors onto position vectors of equal lengths.
Let P be any point. By the assumption, the distance between L(O) and L(P) is equal to the distance between O and P. Therefore |OP'| = |OP| where P' = L(P). Therefore, L preserves lengths of position vectors.

Assume that L preserves norms of all position vectors. To prove that L preserves distance.
Let Q and R be any points, Q' = L(Q) and R' = L(R) their images, with position vectors q, r, q', r' locating them. Let p = r -- q.Then
distance between L(Q) and L(R) squared =

The following provides another condition that a lintr preserves distance.

[2.17] (Preserving distance) A linear transformation preseerves distance if and only if it carries the special triangle onto a congruent triangle.

Assume L preserves distance. To prove the triangles are congruent.
Segments L(O), L(1,0), and L(O), L(0,1) and L(1,0), L(0,1) are congruent respectively to segments O, (1,0), and O, (0,1) and (1,0), (0,1). This fact makes the triangles congruent (sss).

Assume that the triangles are congruent. To prove L preserves distance
This part of the proof uses the special orthonormal vectors i and j. According to [2.16] it will suffice to show that L preserves norms of position vectors,, that is |L(x,y)| = |(x,y)|. As usual it is easier to work with the squares of the norms. Recall that the square of a norm is equal to the inner product of what is between the norm symbols, with itself. The red == in the following chain of equalities is supported by the assumption.

[2.18] (Corollary: preserving congruence of triangles) A linear transformation carries every triangle onto a congruent triangle if and only if it carries the special triangle onto a congruent triangle.

Assume that lintr carries every triangle onto a congruent triangle. Then it carries, in particular, the special triangle onto a congruent triangle.

Assume the lintr L carries the special triangle onto a congruent triangle. To prove that L carries every triangle onto a congruent triangle.
By [2.17] and the assumption, the lintr preserves distance. Then it caries sides (segments) of any triangle onto congruent images of those sides (congruent segments), which could be sides of triangles.

The truth of the following two statements will be assumed.

[2.19a] (Preserving orientation) A distance preserving linear transformation carries a simply closed polygon or circle onto a closed polygon or circle of the same orientation if and only if it carries the special triangle onto a triangle with the same orientation.

[2.19b] (Reversing orientation) A distance preserving linear transformation carries a simply closed polygon or circle onto a closed polygon or circle of the opposite orientation if and only if it carries the special triangle onto a triangle of opposite orientation.

Section 3:   Geometric actions of some linear transformations on the plane

The first and second actions [3.1] and [3.2] introduce these ideas.

[3.1] Action:   Reflection through the origin

[3.1a] (Reflection through the origin) The origin is the midpoint of the segment joining any point and its reflection through the origin.
Notation: if P' is the image obtained by reflecting a point P through the origin, then O is the mid point of segment PP'.

[3.1b] Some immediate results of [3.1a]:
   (i) Vectors OP' = -- OP
   (ii) Only the origin is carried back onto itself (left fixed); for all other points, P' and P are distinct

[3.1c] (Images of the special points and triangle)
  (i) the images of (1,0) and (0,1) are (-1,0) and (0,-1) respectively;
  (ii) the special triangle has vertices O, (1,0), (0,1). Its image has vertices O, (-1,0), (0,-1);
  (c) the orientation of the image triangle is positive (counter-clockwise).

It is obvious that O is the mid point of the segment joining (1,0) and (-1,0)
and the segment joining (0,1) and (0,-1).
The following matrix is formed from the transposes of these images:

[3.1d] (The reflection through the origin is a linear transformation) The linear transformation L defined by

Since the images of (1,0) and (0,1) are (-1,0) and (0,-1) respectively, L(1,0) = (-1,0) and L(0,1) = (0,-1). The equation for L(x,y) is derived from L(x,y) = xL(1,0) + yL(0,1) = x(-1,0) + y(0,-1) = (-x,-y).
To prove that L is the reflection through the origin, see where the mid point of (x,y) and L(x,y) lies. (x,y) + L(x,y) = (x,y) + (-x,y) = (x + (-x), y + (-y)) = (0,0). Therefore (1/2) [(x,y) + (-x,y)] = (0.0). Hence origin O is the mid point of the segment joining (x,y) and L(x,y).

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[3.2] Action:   Similitude, a uniform expansion outward from or a uniform contraction toward the origin

More generally, any positive real number λ can be used in place of 2:

[3.2a] (Similitude) The action of obtaining a point P' from an arbitrary point P by the vector equation (*) is called a similitude with center at the origin and with coefficient λ, a positive real number..

[3.2b] For various values of λ the geometric effects are:
   (i) if λ > 1, then image P' is further from the origin than P. This similitude is a niform dilation of the plane about the origin;
   (ii) if λ = 1, then P'=P and the image of any point coincides with tht point. The similitude is the identity function on the plane, and no dilation nor contraction occurs;
   (iii) if λ < 1 (but λ > 0) then image P' is closer to the origin than P. This similitude is a uniform contraction of the plane about the origin;

A similitude is completely determined by its coefficient. In fact the collection of all similitudes is in one-to-one correspondence with all the positive real numbers. That collection of similitudes is actually a group.
   (i) if one similitude is followed by another the result is a similitude whose coefficient is the product of the two similitudes of the component similitudes;
   (ii) the identity function is a similitude with coefficient 1;
   (iii) the inverse of a similitude is a similitude whose coefficient is the inverse of the first similitude.

[3.2c] (Images of the special points) A similitude with coefficient λ carries the points (1,0) and (0,1) onto the points (&lambda, 0) and (0,λ) respectively.

If E₁ and E₂ are the special points (1,0) and (0,1) spectively, and E'₁ and E'₂ are the images by action of the similitude, then (position vectors)

From the transposes of the images of the special points the following matrix is formed:

[3.2d] (A similitude with center at the origin is a linear transformation) The linear transformation L defined by

Let P(x,y) and P'(λx,λy). Since (λx,λy) = λ(x,y), OP' = λOP. Therefore, L is a similitude with coefficient λ.

[3.2e] (Parallel image) A similitude carries segments onto parallel segments.
Notation: If P' and Q' are images of points P and Q respectively, then segment P'Q' is parallel to segment PQ.

Vectors OP' = λOP and OQ' = λOQ. Converting these vectors to position vectors

[3.2f] (Preserving angles) A similitude carries intersecting lines onto intersecting lines in such a way that the angle of intersection of the second pair of lines is congruent to the angle of intersection of the first pair.
Notation: If points A', B', C' are images of points A, B, C then angle A'B'C' is congruent to angle ABC.

By [[3.2e] segment A'B' and B'C' are parallel to AB and BC respectively. Therefore angle A'B'C' is congruent to angle ABC.

However, similitudes do not preserve lengths of segments. The images of segments are not necessarily congruent to the segments.

[3.3] Action:   Orthogonal reflection of the entire plane about a line through the origin

Upon comparison of the two triangles, two observations can be made.
  (a) the copy on the left side of the paper is congruent to the original triangle on the right side;
  (b) the copy has an orientation opposite to the orientation of the original triangle on the right side.
The vertices of the original triangle on the right side are labeled A,B,C in a counter-clockwise direction. But the labels on the copy A',B',C' are given in a clockwise direction. The process of reflecting the original triangle about the vertical line makes a congruent figure but reverses orientation.

This time a not-yet-dried parallelogram ABDC is located in the upper half of another paper, and a horizontal line is drawn below it. The paper is folded along the horizontal line to produce a copy of the parallelogram A'B'C'D' in the lower half of the paper. (Fig 2). A reflection about the horizontal line has occurred. The parallelograms are congruent but have opposite orientation.

In both examples it is the action of producing the copies that is the focus of these discussions. First is a geometric description of the action that reflects the triangle about the vertical line (Fig 1b). If a horizontal line segment is drawn joining points A and A' (vertices of the triangles) then the vertical line is the perpendicular bisector of segment AA'. Similarly, it is the perpendicular bisector of horizontal segments BB' and CC'. In general, if P is any point, then its image P' determines a horizontal segment PP' of which the vertical line is a perpendicular bisector.

Now to derive the formula for the linear transformation that does the reflecting of the triangle. The lintr requires the environment of the coordinate plane where each point has a label, an array of length 2. Place the original right triangle in the first quadrant of the coordinate plane and the vertical line on the y-axis. Since the vertices are points, they have attached labels, say

[3.3a] (Orthogonal reflection L about the y-axis)
  geometric definition L(P) = P' where the y-axis is the perpendicular bisector of segment PP' or P=P' if point P is on the y-axis.
  algebraic definition L(x,y) = (-x,y).

The algebraic definition makes L a linear transformation, since the expressions inside (-x,y) are linear expressions.

The derivation of the formula for the lintr that reflections the parallelogram about the horizontal line (Fig 2) is obtained in a similar way. But in this case the horizontal line is the perpendicular bisector of vertical segments AA', BB', CC' DD'. Place the original parallelogram in the first quadrant of the coordinate plane and the horizontal line on the x-axis.

[3.3b] (Orthogonal reflection L about the x-axis)
  geometric definition L(P) = P' where the x-axis is the perpendicular bisector of segment PP' or P=P' if point P is on the x-axis.
  algebraic definition L(x,y) = (x,-y).

The algebraic definition makes L a linear transformation, since the expressions inside (x,-y) are linear expressions.
Of course, this lintr is different from the lintr that is a reflection about the y-axis described in [3.1a]..

The reflections about the xaxis and the y-axis are linear transformations. Therefore, they have associated matrices. These matrices are obtained from what happens to the special points (1,0) and (0,1). The reflection about the y-axis carries these points onto (-1,0) and (0,1) respectively. Then the transposes of these images become the columns in the associated matrix.

The reflection about the x-axis carries the special points (1,0) and (0,1) onto (1,0) and (0,-1) respectively. Then the transposes of these images become the columns in the associated matrix.

[3.3c] (Associated matrices for the orthogonal reflections about the axes:

Any orthogonal reflection about a line through the origin must reflect the special points (1,0) and (0,1) about that line. In [3.3d] and [3.3e] that follow assume that the line passes through the origin and has an inclination angle θ.

[3.3d] (Reflected images of the special points) A reflection about the line with inclination angle θ carries (1,0) and (0,1) onto the points

Click here to see a discussion supporting this statement.

[3.3e] (Reflection is a linear transformation) The linear transformation L defined by

Click here to see a discussion supporting this statement.

Statements [3.3d] and [3.3e] combine to support the following:

[3.3f] (Orthogonal reflection about any line through the origin)
  geometric definition If P' is the reflected image of any point P then the line is the perpendicular bisector of segment PP', or P=P' if point P is on the line (Fig 4).
  algebraic definition The matrix associated with the orthogonal reflection about a line through the origin and making an inclination angle θ with the positive x-axis is

The associated matrix comes directly from [3.3d].

If θ = 90° then line_o is the y-axis. The associated matrix (6) becomes the first matrix in (3) above.
If θ = 0° or θ = 180° then line_o is the x-axis. The associated matrix (6) becomes the second matrix in (3) above.
If θ = 45° then line_o bisects the angle between the positive x-axis and the positive y-axis. The reader can replace θ in the formula for L in [3.1e] by 45° (dont forget to double this angle to 90°) and obtain: L(x,y) = (y,x). Therefore L interchanges the coordinates. What is the matrix associated with L ?

[3.4] Action:   Rotation of the entire plane about the origin and through a given angle

A precise geometric definition of a rotation is not given here. It would not be in the spirit of an intuitive introduction to rotations. Instead two equivalent intuitive descriptions [3.4a] and [3.4b] are given below.

[3.4a] (Intuitive geometric description 1) Keep in mind that the rotation function on the plane carries points onto points in a definite manner. The plane does not move, so it will be called the fixed plane. A second plane is invented to carry out the motion of rotation. It is called the plane(See adjacent figure.) It lies above the fixed plane. There is a hole just above the origin in the other fixed plane. An axis is created perpendicular to the fixed plane and runs through the hole.

Suppose this plane is empty. Then each point P in the fixed plane is copied vertically into the movable plane above it to make a point P. After all the points have been copied from the fixed plane, the movable plane is rotated through a given angle. If the angle is positive then the rotation is done counter-clockwise. If the angle is negative then the rotation is clockwise. Each point P is carried along with the rotating movement. After the rotation of the movable plane is completed, then P is above some point P' in the fixed plane. (The new position of the movable plane and point P' are not shown in the figure.)

The rotation function on the fixed plane carries P onto P'.
The phrase "rotation function on a plane" is less common that the phrase "rotation of the plane". The reader may mentally attached the word "function" to the word "rotation" whenever it occurs alone in this section.

The above description places the two parallel planes in three dimentional space. The following is an equivalent description of the rotation function in the two dimentional (fixed) plane, and does not involve any rotating (green) plane. But it does involve an arbitrary point P in the (fixed) plane to be carried onto a point P' in that same plane. It is this description that will lead to a more rigorous algebraic definition of a rotation.

[3.4b] (Intuitive geometric description 2) With one end point fixed on the origin O swing segment origin OP around in a circular motion through angle θ and stop. Then the segment origin OP' is formed. The point P' is the rotated image of point P. If θ is positive, the the motion is counter-clockwise (Fig 1a). If θ is negative then the motion is clockwise (Fig 1b).

[3.4c] There are several immediate results from this description:
   (i) The angle completely determines the rotation.
   (ii) Any point and its rotated image are the same distance from the origin.
   (iii) If θ = 0, there is no motion. Then P' = P. The identity function I is a rotation through an angle of zero.
   (iv) The rotation of any point about the origin through an angle of θ is equivalent to a rotation of that same point through an angle of θ - 360°.
   (v) The rotation through any number of revolutions, positive or negative, carries each point back onto itself.
   (vi) A rotation through an angle -180° is equivalent to a rotation through an angle of 180°.

The proof of the following important statement is almost trivial:

[3.4d] (Combining rotations) If a pair of rotations act on a plane, then the angle of the combined rotations (one acting after the other) is equal to the sum of the angles of the individual rotations.
Notation: If θ₁ and θ₂ are the angles of the two rotations, then their combined action (composition) is a rotation with an angle of θ₁ + θ₂.

The first rotation carries point P onto point P', and the second rotation carries point P' onto P".

The angles involved with rotations are real numbers, which are commutative. This makes the combinedaction of any two rotations commutative - either rotation may act first.
Since θ + -θ = 0, every rotation has an inverse. Simply rotate the same amount but in the opposite direction.

[3.4e] The collection of all rotations about the origin form a commutative group.

It is time now to derive the formula for a rotation. First the images of the special points (1,0) and (0,1) are calculated.

[3.4f] (Rotated images of the special points) A rotation through angle θ carries points (1,0) and (0,1) onto the points

Click here to see the discussion supporting this statement.
From the transpose of coordinates of image points form the matrix

[3.4g] (General formula for rotation) The linear ttransformation L defined by

Click here to see the discussion supporting this statement.

[3.5] Action:   Orthogonal projection of the entire plane onto a line through the origin

[3.5a] (Geometric description) Where the line through the given point intersects the given line perpendicularly is the point called the orthogonal projection of the given point upon the given line. (If the given point is already on the given line then the point is also the orthogonal projection upon the given line.) Do this action for every point in the plane.

Clearly the line segment joining the given point and its projected image is perpendicular to the given line.

[3.5b] Some obvious results of [3.5a]    (i) The location and position of a given line completely determine the orthogonal projection of the entire plane onto it.
   (ii) The image of a line or line segment perpendicular to the given line is a point.    (iii) Parallel line segments not perpendicular to the given line are projected onto congruent segments inside the given line.    (iv) A projection carries subsets of 2 dimensions (have langth and width) onto subsets of lesser dimension (as usbsets of the given line).    (v) There is no inverse action to undo the action of a projection.

Two simple orthogonal projects occur in the coordinate plane if the given lines are the axes. In Fig 3 the arbitrary point P(x,y) is orthogonally projected onto the x-axis, and the arbitrary point Q(x,y) is orthogonally projected onto the y-axis. It is not difficult to determine the image points, namely P'(x,0) and Q'(0,y). (The second coordinate of every point on the x-axis must be 0. The first coordinate of every point on the y-axis must be 0. The projections are vertical and horizontal respectively.)

The vertical projection onto the x-axis of the special points (1,0) and (0,1) is   (1,0) = (1,0) (point is already on the x-axis)   (0,1) ---> (0,0), (point (0,1) is above the origin). Also the horizontal projection onto the y-axis of (1,0) and (0,1) produces images (0,0) and (0,1).

>b>[3.5c] (Images of the special points) The images of (1,0) and (0,1) projected orthogonally onto the x-axis and y-axis are (1,0), (0,0) and (0,0), (0,1) respectively.

The matrices from the transposes of these images are:

[3.5d] (Formula for projection onto the axes) The linear ttransformations L₁ and L₂ defined by

Click here to see an argument supporting this statement.

From the transposes of the images of the special points the following matrix is obtained:

Click here to see an argument supporting this statement.


A projection onto a line is singular because it has no inverse. There is no linear transformation that will undo what a projection has done. This is confirmed by the determinant of (4) which is zero. However, the matrix in (4) has rank 1, making projections onto lines having rank 1. See [2.7] above.

Section 5:   Linear transformations that preserve distance

The most simple of congruent figures are segments. For two segments to coincide it is sufficient that their end points coincide. For a linear transformation L to carry a segment AB onto a congruent segment CD it is sufficient that L(A) = C and L(B) = D. In other words, the distance between points A and B is the same as the distance between points L(A) and L(B).

[5.1] (Preserving distance) A linear transformation preserves distance if and only if for every pair of points the distance between the points is equal to the distance between their images.
Notation: if P and Q are any points then    the distance between P and Q = the distance between L(P) and L(Q).

Another name for a distance preserving linear transformation is a linear isometry. (There are other kinds of isometries. A translation is an isometry which is not a linear translation.) Two figures are congruent if there is an isometry that carries one figure onto the other. Therefore, an isometry may also be called a "congruence transformation" but this name is seldom used.

The linear transformation carries vectors onto vectors. A linear transformation that preserves distance carries vectors onto vectors of equal lengths (norms). This is true of position vectors. If L is a linear transformation that preserves distance, then |L(p| = |p| because distance between O and P   =   distance between O and L(P).

From their introductory intuitive descriptions given in the previous section, it seems obvious that rotations and orthogonal reflections about lines are distance preserving transformations. These facts can be proven by elementary geometry. In the previous section the actions first affected the special points, then extended to all of the plane. That method can be used here. Intuitively speaking, if a linear transformation preserves distance at the origin, then it preserves distance everywhere. Let E₁ and E₂ be the names of the special points (1,0) and (0,1) respectively that is,

Then

This information is more easily stated in terms of the special vectors   i and j. They are position vectors that locate the special points (1,0) and (0,1):

[5.2] A linear transformation preserves distance if and only if it carries the special vectors onto orthonormal position vectors.
Notation: L preserves distance if and only if   L(i) and L(j) are orthonormal position vectors:

Let P be the point with coordinates (x,y). Then position vector p = OP. To show that |L(p)| = |p|. Since   p = (x,y) = x(1,0) + y(0,1)   then   L(p) = xL(1,0) + yL(0,1). By the distance formula, |r|² = x x² + y²
Then using inner products
|L(p)|² =
     (L(p)) * (L(p)) =
     (xL(1,0) + yL(0,1)) * (xL(1,0) + yL(0,1)) =
     x²L(1,0)*L(1,0) + 2xyL(1,0)*L(0,1) + y²L(0,1)*L(0,1) =
     x²|L(1,0)|² + 2xy 0 + y²|L(0,1)|² =
     x² + y² =
     |p|²     &nbs0p;  (by the distance formula)

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[3.1] Action: Reflection of the entire plane through the origin onto that entire plane


  (a) Definition: a function L such that the origin O is always the mid point of the segment joining any point to its image.
        Notation: O is the mid point of segment joining P to P' = L(P).
  (b) Position vectors: L(p) = --p,    p' = --p
  (c) Coordinates: L(x,y) = (--x,--y),    x' = --x, y' = --y
  (d) Distance: Preserved
  (e) Orientation: Preserved
  (f) Matrix: As shown above
  (g) Special Functional Equations: (Involution) L*L = I,    L^-1 = L
  (h) Sets left fixed setwise but not pointwise: All lines through the origin, all circles with centers at the origin, any subset symmetric about the origin
  (i) Points left fixed: origin

The rest of this chapter will make heavy use of angles and trigonometric functions. Click here to review some of them.

Intuitive idea: Keep in mind that the rotation function on the plane carries points onto points in a definite manner. The plane does not move. Therefore a second plane is invented to carry out the motion of rotation.(See adjacent figure.) It lies above the fixed plane. There is a hole just above the origin in the other plane. Suppose this plane is empty. An axis is created perpendicular to the fixed plane and runs through the hole.

Each point P in the fixed plane is copied vertically into the plane above it to make a point P. After all the points have been copied from the plane, the plane is rotated through a given angle. Each point P is carried along with the rotating movement. After the rotation of the green plane is completed, then P is above some point P' in the fixed plane. (The new position of the green plane and point P' are not shown in the figure.)

The rotation function on the fixed plane carries P onto P'.
The phrase "rotation function on a plane" is less common that the phrase "rotation of the plane". In the latter situation the black plane is rotated and the green plane that contains the copies of all points is left fix. However, the black plane is rotated in the opposite direction that was suggested in movement of the green plane described above. After the black plane is rotated, the green point P is copied back down onto the black plane, locating the image point P'. This method of rotation is used only in a few examples. The rotation function, also called rotation, is described in this section. (The reader may mentally attached the word "function" to the word "rotation" whenever it occurs in this section.) The phrase "about the origin" means that the origin is left fixed by the rotation (function).

The vertex of the angle is always at the origin, which is left fixed. Any other point and its image form an isosceles triangle with the angle as sides. The orientation origin-point-image of the triangle is the same as the orientation of the given angle: both are counter-clockwise or both are clockwise.

The orientation requirement distinguishes rotations from another function discussed in [3.3] below.

Let P be any point (except the origin O) and let L be a rotation about the origin. Let θ be the angle of rotation. The initial side of θ passes through P and the terminal side of θ passes through P' = L(P). And |OP| = |OP'| from the definition. The triangle OPP' has the same orientation as θ: both are counter-clockwise (upper figure) or both are clockwise (lower figure).

[3.2a] The signed angle and origin completely determine a rotation.

[3.2b] A rotation about the origin through an angle of zero is equal to the identity function.
Rotating the plane about the origin through angles equal to any number of complete revolutions, positive or negative also produces the identity function.

If θ = 0 then P' = P. SInce P is any point, this means that the rotation leaves every point fixed. This, in turn, means that the rotation is equal to the identity function: L = I
A similar argument exists for θ = 1 rev (360°), 2 ref (720°), ..., - 1 rev, - 2 rev,...

[3.2c] A rotation about the origin through an angle of 1/2 rev (180°). is equal to the reflection through the origin (see [3.1] above).

The initial and terminal sides of an angle of measurement 1/2 rev form a full straight line. The origin is always the mid point of the segment joining any point and its image. But this is the definition of the reflection of points through the origin.

[3.2d] Rotations about the origin preserve distance.

Click here to see a proof of this statement.

[3.2e] Rotations about the origin are linear transformations.

This follows from [3.2d] and [2.2].

[3.2f] Rotations about the origin preserve orientation.

This is from part of the definition of a rotation.

Next is the derivation of the matrix for a rotation L about the origin through an angle θ.

Let L denote a rotation of the plane about the origin through an angle θ. Let

[3.2g] The matrix for a rotation of the plane about the origin through the angle θ is

[3.2h] The equations for finding the coordinates of the image point L(x,y) from (x,y) are

They can be obtained from the matrix product (x',y')^t = M(x,y)^t.

[3.2xx] The product of two rotations (about the origin) is a rotation. Moreover, the angleof the product is equal to the sum of the angles of the two rotations.

Let L₁ and L₂ be any rotations about the origin and let θ₁ and θ₂ be their angles of rotation respectively. SInce the product of linear transformations is a linear transformation, then the product

There exists a line through the origin. The function carries any point onto an image such that the line is a perpendicular bisector of the segment joining the point and its image. If the point is on the line then the image and point coincide.

Let L denote the function and line_o the axis of reflection. Then L reflects the points P,Q,R,W about line_o to produce images P' = L(P), Q' = L(Q), R' = L(R), W' = L(W). The segments PP', QQ', RR' are perpendicular to line_o. Obviously, the location of the axis of reflection completely determines the reflection.

If a mirror is placed vertically on the axis of reflection, then images of things are seen in the mirror. The image of a line segment is a line segment. The line segment and its image seem to have the same length. A proof that the orthogonal reflection about a line preserves distance is based on the fact that

(*)      if there exists a perpendicular bisector to two parallel sides of a trapezoid then the diagonals have the same length, and the other two sides have the same length.

In the adjacent figure ABDC is a trapezoid with parallel sides AB and CD. M,N are mid points of those sides respectively, and MN is the perpendicular bisector of AB and CD. ( MACN is the reflected image of MBDN and conversely.) Click here to see a geometric proof that diagonals |AD| = |BC| and sides |AC| = |BD|. Unproven yet is the fact that AD, BC and MN are concurrent.

Let P and Q be any pair of points. Let P' = L(P) and Q' = L(Q) be their images under an orthogonal reflection L about a line MN. Then MN is the perpendicular bisector of the parallel lines PP' and QQ'. There are two cases. Points P and Q are on the same side of MN (Fig 1a) and P and Q are on opposite sides of MN (Fig 1b). Both cases satisfy the conditions of the above theorem (*). The darkened sides PQ and P'Q' in Fig 1a have the same length. The darkened diagonals PQ and P'Q' in Fig 1b have the same length. Therefore, orthogonal reflection L preserves distance.

[3.3a] (Distance preserving) Every orthogonal reflection about a line preserves distance.

By [2.2]

[3.3b] (Linear transformation) Every orthogonal reflection about a line through the origin is a linear transformation.

The word "orthogonal" means perpendicular. It is possible to define a "slant reflection about an axis" in which the segtments joining points to their images are all parallel, and the axis passes through the midpoints of those segments but is not perpendicular to those segments. The reflection is a linear transformation but it does not preserve all distances.

It is convenient to place the reflection in the familiar x,y-coordinate system. Let L be the orthogonal reflection about a line_o through the origin making an inclination angle θ with the positive x axis. The angle is measured positively, meaning counterclockwise. That angle determines completely the location of the line_o.

If 0° < θ < 90° then the line_o goes from lower left to upper right (Fig 2a). It is in quadrants I and III. But if 90° < θ < 180° then the line_o goes from lower right to upper left (Fig 2b). It is in quadrants II and IV. There is no need for    θ >= 180° because in that situation θ and θ -- 180° locate the same line_o.

L is a linear transformation and therefore its actions are completely determined by what it does to the special points

Since orthogonal reflection preserves angles,

The equations

The adjacent figure shows that a reflection about a line reverses orientation. In that figure the labels L(A), L(B), L(C) and the labels A, B, C are in opposite order. It is not a coincidence that the determinant of the above matrix (8) associated with the reflection is negative (more exactly -1).
If the line is the edge view of a vertical mirror, the image in the mirror of triangle ABC would be triangle L(A)L(B)L(C) with its reverse orientation. Hence the name "reflection" for the function L.

Click here for a more direct derivation of equations (9).

Section 4:   Distance preserving

Section 2:   Linear transformations that preserve distance

If a function preserves distance then it carries any geometric figure (or any subset) in the plane onto an image without distortion. The figure and image are congruent. For example, the function carries segments onto congruent segments, triangles onto congruent triangles, circles onto congruent circles, picture of a house onto an exact copy of that picture,... The condition that two figures be congruent is that there be a distance preserving function carrying one figure onto the other. A distance preserving function may be described as a rigid motion. That function picks up a figure (or subset) and carries it without "damage" onto a congruent figure (or subset). However, it is possible that during the motion the figure is flipped over.

What distance preserving functions do with the origin is important. It is possible to produce two simple congruent figures, but no distance preserving linear tranformation can carry the first figure onto the second. On the horizontal x-axis, the segment with end points (0,0) and (1,0) is congruent to the segment with end points (2,0) and (3,0). Any function that carries the first segment into the second must carry the origin onto some point in the second segment. But by [1.1] this is enough to prove that no linear transformation may carry the first segment onto the second. (The situation will be resolved in the next chapter with the introduction of a new kind of geometric function that is affine.)

[2.2] Any distance preserving function on the plane that carries the origin onto itself is a linear transformation.

The proof of [2.2] is a result of two theorems proven in the next chapter 7 on affine transformations: (a) Every distance preserving function on the plane is affine, and (b) every affine transformation that carries the origin onto itself is linear.

This theorem [2.2] means that some congruences of plane geometry may be described by distance preserving linear transformations. However, because of [1.1] the origin must be located appropriately. For example, rotations must be about the origin, For reflections about lines, the lines must pass through the origin. For a reflection through a point, the point must be the origin.

Suppose that L is a linear transformation that preserves distance. If C is a circle then the image L(C) is a congruent circle. And if the circle C has center at the origin, the L(C) has its center at the origin. This means that C and L(C) coincide and are the same circle. If P is any point on C the L(P) is also on C. Therefore, the segment from O to P and the segment from O to L(P) have the same length. This means that the position vectors p and L(p) have equal norms:

[2.3a] (Preserving norms) A linear transformation preserves norms if and only if the norm of any vector is equal to the norm of the image of that vector.
Notation: L preserves norm   if and only if   |L(p)| = |p| for any position vector p.

Preserving norms means that any point and its image are the same distance from the origin.

Example 1: the linear transformation L defined by L(x,y) = (y,-x) preserves norms because

Example 2: the linear transformation L defined by L(x,y) = (x sin θ + y cos θ , -x cos θ + y sin θ) preserves normes because |L(x,y)|² = |(x sin θ + y cos θ , -x cos θ + y sin θ)|²
              = (x sin θ + y cos θ)² + ( -x cos θ + y sin θ)²
              = x² sin²θ + 2xy sin θ cos θ + y cos²θ   +   x² cos²θ - 2xy sin θ cos θ + y sin²θ
              = x² (sin²θ + cos²θ) + x² (cos²θ + sin²θ)
              = x² + y²
              = |(x,y)|²
Therefore, |L(x,y)| = |(x,y)| which shows that L preserves norms.

[2.3b] (Preserving distance and norms) A linear transformation preserves distance if and only if it preserves norms
Notation: L preserves distance   if and only if   |L(p)| = |p| for all position vectors p.

If a function preserves distance, then it preserves the distance between the origin and points.
Conversely, suppose L preserves norms. Let Q and R be any points. Let p = q -- r.

If a linear transformations preserves distance between the origin and any point, then it will preserve the distance between any point and any other point. This indicates the nature of linear transformations. Intuitively speaking, if they do something at the origin then they do the same thing at any other point. This generalization is a direct result of the fact that every linear transformation is a homomorphism.

Click here to see a discussion describing and supporting the following two statements.

[2.4] (Preserving angles) If a linear transformation preserves distance then it preserves angles.
Notation: if L preserves distance and if θ is the angle between line₁ and line2 then θ is the corresponding angle between L(line₁) and L(line₂).

Because of similarity between figures of different sizes, preserving angles is not a strong enough condition to guarantee a preservation of distance everywhere.

[2.5] (Preserving inner product) A linear transformation preserves distance if and only if it preserves inner product.
Notation: L preserves distance   if and only if   L(p)*L(q) = p*q for all position vectors p and q.

Section 5:   Fixed points

[2.8] (Fixed points) A linear transformation on the plane leaves a point fixed if the linear transformation carries that point back onto itself.
Notation: P is a fixed point of a linear transformation L if and only if   L(P) = P.

By [2.1] every lintr on the plane has at least one fixed point, the origin. There may or may not be more fixed points. Since the identity transformation I on the plane carries each point of the plane onto itself, every point of the plane is a fixed point of the identity transformation.

What are the fixed points of the lintr L defined by  

[1.4] (Line of fixed points) If there are two distinct fixed points of a linear transformation on the plane, then every point on the line passing through those two points is also a fixed point. Of interest and sometimes of importance are sets left fixed setwise and sets fixed pointwise by the linear transformations. The identity transformation I leaves every point fixed. Every linear transformation L leave the origin fixed: L(O) = O.