Show that there isn't $A \in M_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}$?

Solution 1:

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.

Solution 2:

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.