10
$\begingroup$

What's the easiest and simplest way to show that there isn't $A \in M_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}$?

The nicest solution will award its writer with 500 points.

Thank you

  • 0
    *Le* \*sigh\*. Just so we're above the board: I was laughing at the different ways one could interpret Georges's original comment (and certainly not at Georges or the OP). I thought it was patently obvious that I knew Georges meant the sense of "this problem is seven years old". Oh well...2011-08-21

3 Answers 3

20

Suppose $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}.$

Then $B=A^{1002}$ is a real 2x2 matrix such that $B^{2}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}.$ We show that this is not possible.

Let $\quad B=\begin{pmatrix} a &b \\ c&d \end{pmatrix}.$

Then $B^{2}=\begin{pmatrix} a^{2}+bc &ab+bd \\ ac+cd&d^{2}+bc \end{pmatrix}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}\quad \quad \quad (1)$

Considering the (1,2) entry in equation (1) yields $ab+bd=0.$If $b \neq 0 $ then we must have $a=-d$ but then the diagonal entries of $B^{2}$ would be equal contradicting (1). If $b=0$ then by considering the (1,1) entry in equation (1) we see that $a^{2}=-1$, which is not possible for $a$ real.

  • 0
    Dear @Peter: thank you for the answer, when I'll be able to start a bounty and award you, I'll do that. I believe it will be available tomorrow.2011-08-19
12

If $\lambda$ and $\mu$ are the eigenvalues of $A$ then $\lambda^{2004}=-1$ and $\mu^{2004}=-2$. This implies that $\lambda$ and $\mu$ are complex, non-conjugate eigenvalues of $A$. But for any real matrix all non-real eigenvalues come in conjugated pairs.

  • 0
    As Aaron indicates I am not assuming that $A$ is diagonal. To elaborate somewhat any matrix $A$ can be written as $S^{-1}TS$ where $T$ is a triangular matrix and $S$ is some invertible matrix. (There are many ways of proving this and you may have seen a proof under the name of the Jordan normal form, or the Schur normal form, which are particular ways of writing $A$ in such a way.) The diagonal elements of $T$ are the eigenvalues of $A$. Furthermore $A^{2004}=S^{-1}T^{2004}S$.(Try a smaller power than 2004 to be convinced of this.) Now see what the diagonal elements of $T^{2004}$ are.2011-08-19
7

We assume that such a $A$ exists. We have $A^2-\mathrm{Tr}(A)A +(\det A)I_2= 0$ hence $A^{2004}$ can be written as $\alpha I_2+\beta A$ where $\alpha$ and $\beta$ are real, since $\mathrm{Tr}(A)$ and $\det A$ are real (and $\beta\neq 0$ because $A^{2004}\neq \alpha I_2$). Since $\beta A =A^{2004}-\alpha I_2$, $A$ is diagonal with real entries, namely $A= \begin{pmatrix}d_1&0\\0&d_2\end{pmatrix}$ but since $d_1$ is real we can have $d_1^{2004}=-1$. We can take $2p$ where $p\in\mathbb N$ instead of $2004$.

  • 4
    @Aaron: Yes, but the general statement of the C-H theorem is not needed here because the 2-by-2 case can be verified manually. I think this proof has a certain "wow" factor to the novices. Those proofs that make straightforward uses of eigenvalues are just a bit too boring.2011-08-19