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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)

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These matrices are small. Pick an arbitrary $M$ and compute $M^{-1}BM$, where $B$ is the matrix above. You shouldn't get something too messy. Then make observations about the matrix you get. –  Isaac Solomon Mar 28 '13 at 17:43
Well, the matrix has to have eigenvalues $1,-1$ so the characteristic polynomial $x^2-(a+d)x + (ad-bc)$ must be $x^2-1$ - $a+d=0$ and $ad-bc=-1$. –  Thomas Andrews Mar 28 '13 at 17:45

3 Answers 3

up vote 5 down vote accepted

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 $M$.

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Actually I don't think you have to do all that work (unless you actually need $M$). If $d = -a$ and $a^2 + bc = 1$, then the matrix has the right trace and determinant, and since it is 2-by-2, its eigenvalues are determined by its trace and determinant and are $1$ and $-1$. It is standard theory that the matrix is diagonalizable (we have done this already in our class), and the diagonal matrix in the middle is the matrix given in the question. Then use the definition of "similar". Your answer, which I accepted, has the advantage of constructing $M$. –  Stefan Smith Mar 28 '13 at 20:29
@StefanSmith Yeah, I just wanted to show it explicitly. –  Thomas Andrews Mar 28 '13 at 20:31

Make a picture, your matrix mirrors the $e_2$ vector and doesn't change anything at the $e_1$ vector. The matrix is in the orthogonal group but not in the special orthogonal group. Show that every matrix $$\begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ \sin(\alpha) & -\cos(\alpha)\\ \end{pmatrix} $$ make the same.

Those are the nicest matrix which can happen to you but there are some more (those matrices appear when $M$ itself is in the orthogonal group.

When $M$ is not in the orthogonal group it still won't change the eigenvalues (I am not sure if you already know waht eigenvalues are), $\lambda$ is an eigenvalue to a vector $v\neq 0$ if $$ A \cdot v=\lambda v$$ which means the vector is only enlarged or made smaller through the matrix, but not rotated or something like that. As $A$ has the eigenvalues $1$ and $-1$ you will always find vectors $v_1,v_2$ such that $$ B \cdot v_1= v_1$$ and $$ B\cdot v_2= -v_2.$$

So those matrices won't change one vector and the other one is "turned around".

The eigenvectors of the matrix: $$ \begin{pmatrix} a & b \\c & d \\ \end{pmatrix}^{-1}\cdot \begin{pmatrix} 1 & 0 \\ 0 & -1\\ \end{pmatrix} \cdot \begin{pmatrix} a & b \\c & d \\ \end{pmatrix}$$ are $$\begin{pmatrix} \frac{a}{c} \\ 1 \end{pmatrix} \qquad \begin{pmatrix} \frac{b}{d} \\1 \end{pmatrix} $$ when $c$ and $d$ are not zero,

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that's only some of them - you've only taken orthogonal $M$. –  Thomas Andrews Mar 28 '13 at 17:49
@ThomasAndrews fixed it –  Dominic Michaelis Mar 28 '13 at 17:56

The question says "describe in words", so I think an answer without much formulas is in place. Conjugates of this matrix will be precisely those $2\times2$ matrices with eigenvalues $+1$ and $-1$, in other words those which fix the scalar multiples of one nonzero vector, and negates the scalar multiples of another (necessarily linearly independent) vector. Another description is any matrix of an involution (i.e., with square the identity) other that plus or minus the identity (since their minimal polynomial $X^2-1$ is split with simple roots, involutions are always diagonalisable (except in characteristic $2$)).

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