I will cover this first since this was a topic that was omitted from some versions of the textbook.

**Invariant Points**

Let T be a linear transformation. We say that a point \mathbf x is **invariant** under T if T(\mathbf x) = \mathbf x.

For example, if T is the identity transformation in \mathbb R^2 represented by the matrix \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, every point (x,y) \in \mathbb R^2 is invariant, since T(x,y) = (x,y) for all x, y. It’s actually quite straightforward to find **all** invariant points of a given transformation.

Take for example the transformation T represented by the matrix \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}. We know that (x,y) is an invariant point of T if and only if:

T \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}x \\ y\end{pmatrix}

That is:

\begin{pmatrix}2x + y \\ x + 2y\end{pmatrix} = \begin{pmatrix}x \\ y\end{pmatrix}

so we have 2x + y = x and x + 2y = y, which gives x + y = 0 in both cases.

So any point on the line x + y = 0 is invariant under T, and in fact any invariant point of T lies on x + y = 0. We say that this is a **line of invariant points**, not to be confused with the next definition.

**Invariant Lines**

We say that a line L is **invariant** under T if every point on L is mapped onto another point on L. Note that since each point on a line of invariant points is mapped to itself by T, any line of invariant points is an invariant line. However - an invariant line is not necessarily a line of invariant points. A point on an invariant line may be mapped onto **any other point on the line** by T, not *necessarily* itself.

As an example, we will return to the transformation T represented by the matrix \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix} and find its invariant lines.

Say y = mx + c is an invariant line of T (we know that there’s at least one). We know that each point of the form (x, mx + c) will be mapped to another point of the same form. To write this more usefully:

\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}x \\ mx + c\end{pmatrix} = \begin{pmatrix}X \\ mX + c\end{pmatrix}

for some real number X.

This gives:

\begin{pmatrix}2x + mx + c \\ x + 2mx + 2c\end{pmatrix} = \begin{pmatrix}X \\ mX + c\end{pmatrix}

so X = 2x + mx + c = (m + 2)x + c and x + 2mx + 2c = (2m + 1)x + 2c = mX + c, so (2m + 1)x + c = mX.

We can substitute the former into the latter to give:

m((m + 2)x + c) = (2m + 1)x + c

Expanding everything out:

m^2 x + 2mx + mc = 2mx + x + c

rearranging and factoring you get:

(m^2 - 1)x + c(m - 1) = 0

Since this must hold for *all* x \in \mathbb R, we must have m^2 - 1 = 0 and c(m - 1) = 0. From the former we find that either m = 1 or m = -1. If m = 1, then c(m - 1) = 0 for all c, so any c works. If m = -1, we must have -2c = 0, so c = 0.

So the invariant lines of T are the lines of the form y = x + c with c \in \mathbb R, and the line y = -x. (which we found in investigating invariant points)