1. ## Basis question

For the given matrix H in M2(R) and for any real number r let Vsubr denote the solution space in M2(R) of the equation

[xy]
[zw]H =

rH[xy]
[zw]

Find a basis for Vsub2 given that
H = [01]
[-10]
Okay. So I had some difficulty typing this question so it looks really awkward. But I think you can figure out what I'm doing. H is a 2X2 matrix.
[xy]
[zw] is supposed to be a 2x2 matrix also.

thanks.

2. I think what you are saying is this: Let $V_r$ be the subspace of 2 by 2 matrices satisfying
$\begin{bmatrix}x & y \\ z & w\end{bmatrix}\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}= r\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x & y \\ z & w\end{bmatrix}$.

Find a basis for that space.

Okay, go ahead and do the computation: this says
$\begin{bmatrix}-y & x \\ -w & z\end{bmatrix}= \begin{bmatrix}rz & rw \\ -rx & -rw\end{bmatrix}$.

And that reduces to 4 equations: -y= rz, x= rw, -w= -rx, and z= -ry. Since two involve only y and z and two only x and w, that is really two sets of twe equations. Divide the first equation by the last to get -y/z= -z/y which is equivalent to $y^2= z^2$. That means we must have either y= z or y= -z. But what about r? If y= z, then the first equation above becomes -y= rz so r must be -1. If y= -z, then r must be 1.
Divide the second equation the third to get -x/w= -w/x or $x^2= w^2$. We must have either x= w or x= -w. Again, if x= w then r must be -1 and if x= -w, r= 1.

That is, r must be either 1 or -1. Then matrices in $V_{1}$ are of the form $\begin{bmatrix}x & y \\ -y & -x\end{bmatrix}$ $= x\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\-1 & 0\end{bmatrix}$ so a basis for this space is the two matrices $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ and $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$.

Matrices in $V_{-1}$ are of the form $\begin{bmatrix} x & y \\ y & x\end{bmatrix}$ $= x\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+ y\begin{bmatrix}0 & 1 \\ 1& 0\end{bmatrix}$ so that a basis for $V_{-1}$ is given by the two matrices $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$

3. Yes, but here's the problem. He wants me to find a basis for Vsub2. But you're saying that's impossible. Maybe there is a problem with his question?