The Problem:
Suppose $A$ represents a real $n \times n$ matrix satisfying $A^6 = -A^2$.
(a) Prove that if $A$ is symmetric, then $A = 0$.
(b) Prove that if $n$ is odd, then $A$ is not invertible.
(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).
My Approach:
For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $\lambda_1, ..., \lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $\lambda_1, ..., \lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $\lambda_i^6 = -(\lambda_i^2)$, for each $i \in \{1,2,...,n\}$. But since $\lambda_1, ..., \lambda_n \in \mathbb{R}$, it must then be that $\lambda_1 = \cdots =\lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?
For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $\text{det}(A)^6 = (-1)^n\text{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $\text{det}(A)$ must be $0$; and so $A$ is not invertible.
For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g., \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $\mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2\iff z=0\vee z^4=-1.$$So, take a fourth root of $-1$, such as $\cos\left(\frac\pi4\right)+i\sin\left(\frac\pi4\right)$. Now, write $z$ as a $2\times2$ matrix:$$z=\begin{bmatrix}\frac{\sqrt2}2&-\frac{\sqrt2}2\\\frac{\sqrt2}2&\frac{\sqrt2}2\end{bmatrix}.$$This approach works because$$\begin{array}{ccc}\mathbb C&\longrightarrow&\mathbb{R}^{2\times2}\\a+bi&\mapsto&\begin{bmatrix}a&-b\\b&a\end{bmatrix}\end{array}$$is an injective ring homomorphism.