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

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