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Show that every $2\times 2$ matrix $A$, for which $A^2=-I$, is similar to $\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$.

I need help proving this. Don't have any idea how to proceed. Obviously look at the matrix as a linear operator...

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    $\begingroup$ This is true if $A$ is real and is viewed as a $\mathbb{R}$-linear operator $A:\mathbb{R}^2\longrightarrow \mathbb{R}^2$. In that case, $A$ has no eigenvalue, so for any $x\neq 0$, $x$ and $Ax$ are linearly independent, and so $(x,Ax)$ is a basis of $\mathbb{R}^2$. What is the matrix of $A$ in a basis of the form $(x,Ax)$? $\endgroup$ – Julien May 5 '13 at 14:30
  • $\begingroup$ Why A has no eigenvalues? $\endgroup$ – Jim May 5 '13 at 14:39
  • $\begingroup$ Because an eigenvalue would be a root of $X^2+1$. No such things in $\mathbb{R}$. $\endgroup$ – Julien May 5 '13 at 14:42
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    $\begingroup$ I upvoted your answer but OP might need a reminder that two 2-by-2 matrices with the same eigenvalues, which are distinct, must be similar. This is tricky and s/he might not have been shown this well. $\endgroup$ – Stefan Smith May 6 '13 at 0:06
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Your statement is wrong. Let $A=\pmatrix{\pm i&0\\0&\pm i}$, there are at least two different forms. It would be true if you assume $A$ is a real matrix, in that case the eigenvalues of $A$ should be different, obtaining $i$ and $-i$, the statement follows.

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  • $\begingroup$ Why was this answer downvoted? +1. $\endgroup$ – Hu Zhengtang May 5 '13 at 14:11
  • $\begingroup$ Still no clue ... Let A be real then. $\endgroup$ – Jim May 5 '13 at 14:22
  • $\begingroup$ I upvoted your answer but OP might need a reminder that two 2-by-2 matrices with the same eigenvalues, which are distinct, must be similar. This is tricky and s/he might not have been shown this well. $\endgroup$ – Stefan Smith May 6 '13 at 0:07
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Proof over the real matrices:

Let $A=\pmatrix{a&b\\c&d}$ be a real matrix such that $A^2=-I$. Then

$$A^2=\pmatrix{a&b\\c&d}\pmatrix{a&b\\c&d}=\pmatrix{a^2+bc&b(a+d)\\c(a+d)&d^2+bc}=\pmatrix{-1&0\\0&-1}=-I$$

and

$\left\{\begin{array}{l l} a^2+bc=-1\\ b(a+d)=0\\ c(a+d)=0\\ d^2+bc=-1 \end{array}\right.$

First off, we see that $b \neq 0 \neq c$, since $x^2=-1$ has no real solution. Hence $d=-a$ and $c=-\frac{1+a^2}{b}$, so

$$A=\pmatrix{a&b\\-\tfrac{1+a^2}{b}&-a}$$

Furthermore,

$$\pmatrix{1&-\tfrac{1}{a}\\-\tfrac{1+a^2}{ab}&0}^{-1}\pmatrix{a&b\\-\tfrac{1+a^2}{b}&-a}\pmatrix{1&-\tfrac{1}{a}\\-\tfrac{1+a^2}{ab}&0}=\pmatrix{0&-1\\1&0}$$

Proof over the complex matrices:

It can be shown that every complex (including real) matrix $A$ is similar to its Jordan normal form $J=P^{-1}AP$, where $P$ is a certain invertible matrix. Since $A^2=-I$, it follows that

$$J^2=(P^{-1}A\underbrace{P)(P^{-1}}_{I}AP)=P^{-1}\underbrace{A^2}_{-I}P=-P^{-1}P=-I$$

Now, we write $J=D+M$, where $D$ is a diagonal matrix and $M^2=0$. Then

$$J^2=(D+M)^2=D^2+DM+\underbrace{MD}_{DM}+\underbrace{M^2}_{0}=D^2+2DM$$

But $J^2=-I$ is diagonal, so we must have that $M=0$. Hence $J$ is also diagonal and

$$J=\pmatrix{\pm i&0\\0&\pm i}$$

Since we want $J$ to be similar to a real matrix, the diagonal elements must have opposite signs, so that the sum of the eigenvalues is real. Clearly, both possible matrices are similar to $\pmatrix{0&-1\\1&0}$. For example,

$$\pmatrix{i&1\\\tfrac{1}{2}&\tfrac{i}{2}}^{-1}\pmatrix{-i&0\\0&i}\pmatrix{i&1\\\tfrac{1}{2}&\tfrac{i}{2}}=\pmatrix{0&-1\\1&0}$$

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    $\begingroup$ I think that if the question is about similarity of real matrices to a rotation matrix, the most natural interpretation is that one should show that they are similar over the real numbers. In the setting of this proof that means showing that $P$ can be taken to be a real matrix. $\endgroup$ – Marc van Leeuwen May 5 '13 at 16:06
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    $\begingroup$ How did you figure out P matrix for real matrices? $\endgroup$ – Matej May 7 '13 at 21:51
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    $\begingroup$ @David I set $$P=\pmatrix{1&A\\B&0}$$ and then solved the equation $$\pmatrix{a&b\\-\tfrac{1+a^2}{b}&-a}P=P\pmatrix{0&-1\\1&0}$$ for $A$ and $B$. $\endgroup$ – Librecoin May 9 '13 at 9:21
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    $\begingroup$ @Tharsis One more thing, why did you wrote $P$ as $P=\pmatrix{1&A\\B&0}$? $\endgroup$ – Matej May 9 '13 at 18:08
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    $\begingroup$ @David It was an ansatz; it seemed like two degrees of freedom would suffice. Naturally, to simplify as much as possible, I first tried $$P=\pmatrix{0&A\\B&0},$$ but this implied that $A=B=0$. Seeing where it went wrong, I chose $$P=\pmatrix{1&A\\B&0}$$ instead. $\endgroup$ – Librecoin May 9 '13 at 18:27

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