Let $A$ be a symmetric $n \times n$ matrix. Prove that $A$ and $A^5$ have the same null space.
This question came up when I was solving. And I'm not entirely sure on how to approach it, is it valid to just write $A^4(AX) = 0$ (assuming here that $X$ belongs to the null space of $A$), which would give $0$, meaning that $A$ and $A^5$ have the same null space?
We clearly have $\ker A\subset \ker A^5\subset \ker A^8.$ Assume $v\in \ker A^8.$ Then by symmetry we have $$0=\langle A^8v,v\rangle =\langle A^4v,A^4v\rangle $$ Hence $A^4v=0,$ i.e. $v\in\ker A^4.$ Applying the same procedure twice we get $v\in \ker A.$ Therefore $\ker A^5\subset \ker A^8\subset \ker A.$
The key is that real symmetric matrices have a spanning set of eigenvectors (i.e. the set of its eigenvectors spans $\mathbb{R}^n$), thus assume the vectors $(\mathbf{v}_i)$ form a basis for $\mathbb{R}^n$, where each $\mathbf{v}_i$ is an eigenvector of $A$.
Then to show, that the nullspace of $A^5$ is included in the nullspace of $A$ we need to show
$$ A^5 \mathbf{v} = 0 \implies A\mathbf{v}=0$$
Note that we can write $\mathbf{v}$ uniquely as a linear combination of eigenvectors $(\mathbf{v}_i)$ as
$$ \mathbf{v} = \sum \alpha_i \mathbf{v}_i $$
But then
$$ A^5\mathbf{v} = \sum \lambda_i^5 \alpha_i \mathbf{v}_i = 0$$
Which implies
$$ \lambda_i^5 \alpha_i = 0 $$
Which in turn implies
$$ \lambda_i \alpha_i = 0$$
But this means the only non-zero coefficients $\alpha_i$ were paired with eigenvectors that have a $0$ eigenvalue, i.e. vectors in the nullspace of $A$. So $\mathbf{v}$ is a linear combination of vectors in the nullspace of $A$, hence $\mathbf{v}$ is in the nullspace of $A$, completing the proof.
