Math people:
I assigned this problem as homework to my students (from Strang's "Linear Algebra and its Applications", 4th edition):
Describe in words all matrices that are similar to $$\begin{bmatrix}1& 0\\ 0& -1\end{bmatrix}$$ and find two of them.
Square matrices $A$ and $B$ are defined to be "similar" if there exists square invertible $M$ with $A = M^{-1}BM$ (or vice versa, since this is an equivalence relation). The answer to the problem is not in the text, and I am embarrassed to admit I am having trouble solving it. The problem looked easy when I first saw it.
The given matrix induces a reflection in the $x_2$-coordinate, but I don't see how the geometry helps. A similar matrix has to have the same eigenvalues, trace, and determinant, so its trace is $0$ and its determinant is $-1$. I spent a fair amount of time on it, with little progress, and I can spend my time more productively. This problem is #2 in the problem set, which suggests that maybe there is an easy solution.
I would settle for a hint that leads me to a solution.
EDIT: Thanks to Thomas (?) for rendering my matrix in $\LaTeX$.
Stefan (STack Exchange FAN)
Best Answer
If the matrix is $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ then you know that $a+d=0$ and $ad-bc=-1$. So $d=-a$, and we have that $-a^2-bc = -1$ or $a^2+bc=1$.
This is exactly all of them. You just need to find the eigenvectors for these matrices to find the $M$.
If $a\neq 1$ and $a^2+bc=1$ then we can set $$M^{-1}=\begin{bmatrix}b&a-1\\1-a&c\end{bmatrix}$$
Then $$M=\frac 1{2-2a}\begin{bmatrix}c&a-1\\1-a&b\end{bmatrix}$$
Now just do the calculation to show that the matrix $$M^{-1}\begin{bmatrix}1&0\\0&-1\end{bmatrix}M =\begin{bmatrix}a&b\\c&-a\end{bmatrix}$$
When $a=1$, you have to use a different $M$.