# What does the given fact that a matrix A times itself equals -A have to say about its eigenvalues?

My answer to a question on Quora: What does the given fact that a matrix A times itself equals -A have to say about its eigenvalues?

There were two answers in this thread, and both correctly stated that the eigenvalues of $$A$$ equal $$0$$ or $$-1$$. However, one of the answers contains the assertion:

They are all either zero or $$-1$$, regardless of $$A$$’s being diagonalizable or invertible.

I wish to add that $$A$$ is actually diagonalisable. And here is a proof.

Let $$A$$ be a $$n\times n$$ matrix. To prove that $$A$$ is diagonalisable, it suffices to show that the vector space $$V= \mathbb{R}^n$$ can be written as a sum of eigenspaces $$V_0$$ and $$V_{-1}$$ for $$A$$ and its eigenvalues $$0$$ and $$-1$$, correspondingly, or, which is the same, every vector $$\vec{v} \in V$$ can be presented as a sum

$$\vec{v} = \vec{v}_0 + \vec{v}_{-1}$$

with $$\vec{v}_0\in V_0$$ and $$\vec{v}_{-1}\in V_{-1}$$. But this is easy; indeed, obviously

$$\vec{v} = (\vec{v} + A\vec{v}) – A\vec{v}.$$

Observe that

$$A(\vec{v} + A\vec{v}) = A\vec{v} +A^2\vec{v} = A\vec{v} – A\vec{v} = 0$$

and

$$A(-A\vec{v}) = -A^2\vec{v} = -(-A\vec{v}),$$

so we can take

$$\vec{v}_0 = \vec{v} + A\vec{v} \quad \mbox{ and }\quad \vec{v}_{-1} = -A \vec{v}.$$