Vtyjkyuk

Every time I try a question on this topic I get it wrong.

My textbook says:

Invariant points satisfy $Bbeginpmatrixu\ vendpmatrix=beginpmatrixu\ vendpmatrix$

Re-write this as a system of equations.

Check whether both equations are in fact the same.

If so, they give a line of invariant points.

So I tried this question using that method (and the method needed for finding invariant lines):

For [this] matrix, find any lines of invariant points and any other invariant lines through the origin.

$beginpmatrix3&-2\ 4&-3endpmatrix$

What I did was:
$beginpmatrix3&-2\ 4&-3endpmatrixbeginpmatrixu\ muendpmatrix=beginpmatrixu\ vendpmatrix=beginpmatrix3u-2mu\ 4u-3muendpmatrix$

These equations are not the same, so no lines of invariant points.

Then, to find invariant lines:

$beginpmatrix3&-2\ 4&-3endpmatrixbeginpmatrixu\ muendpmatrix=beginpmatrixu'\ v'endpmatrix=beginpmatrix3u-2mu\ 4u-3muendpmatrix$

$4u-3mu=mleft(3u-2muright)$

$2m^2-6m+4:=0:Rightarrow :m:=1: $or$ :m=2$

So the invariant lines should be $y=2x$ or $y=x$. As far as I know this is sort of right, except $y=x$ is apparently also a line of invariant points. What did I do wrong?

Let
$$mathbfB = left [ beginmatrix b_11 & b_12 \ b_21 & b_22 endmatrix right ], quad mathbfp = left [ beginmatrix u \ v endmatrix right ]$$
If $mathbfp$ is an invariant point with respect to $mathbfB$, then
$$mathbfB mathbfp = mathbfp tag1labelNA1$$
which is equivalent to
$$leftlbrace beginaligned
b_11 u + b_12 v &= u \
b_21 u + b_22 v &= v endaligned right .$$
and
$$leftlbrace beginaligned
(b_11 - 1) u + b_12 v &= 0 \
b_21 u + (b_22 - 1) v &= 0 endaligned right . tag2labelNA2$$
To find invariant points, you solve $eqrefNA2$ for $u$ and $v$. In some cases, the solution is not a single point, but a line. If the matrix is all zeros, then all points are invariant.

Let's examine $mathbfB = left [ beginmatrix 1 & 4 \ 2 & -1 endmatrix right ]$. Substituting to $eqrefNA2$ we get
$$leftlbracebeginaligned
(1 - 1)u + 4 v &= 0 \
2 u + (-1 - 1) v &= 0 \
endaligned right . quad iff quad leftlbracebeginaligned
4 v &= 0 \
2 u - 2 v &= 0 \
endalignedright .$$
The only solution to this is $u = v = 0$. So, this matrix has an invariant point at origin.

Let's examine $mathbfB = left [ beginmatrix 3 & -2 \ 4 & -3 endmatrix right ]$. Substituting to $eqrefNA2$ we get
$$leftlbracebeginaligned
2 u - 2 v &= 0 \
4 u - 4 v &= 0 \
endalignedright .$$
If we divide the upper one by 2, or the lower one by 4, we get the same equation: $u - v = 0$. Thus, this matrix has an invariant line $u = v$.