is $\alpha^2$ positive definite, as $\alpha$ is positive definite

Question

Let $V$ be an inner product space and let $\alpha \in End(V )$ be positive definite. Is $α^2$ necessarily positive definite?

I can say that $\alpha^3$ is positive definite, as $\alpha$ is positive definite, but I have no idea how to show $\alpha^2$ is positive definte or if it's not, nothing came to my mind as a counterexample! I appreciate any help.

As, $\alpha$ is positive definite, so it's self-adjoint and for any nonzero vector $v\in V$, we have $\langle \alpha(v),v\rangle>0$.

Since $\alpha^3$ is self-adjoint and for any given $0\neq v\in V$, we have $$\langle\alpha^3(v),v\rangle=\langle\alpha^2(v),\alpha(v)\rangle=\langle\alpha(\alpha(v)),\alpha(v)\rangle>0$$

Answer 1

$( \alpha^2(v), v) = (\alpha(v) , \alpha(v)) = ||\alpha(v)||^2 > 0$, as $\alpha$ has no kernel, so $\alpha(v) \not = 0$. We use that $\alpha$ is self adjoint in the computation.

Moreover, $\alpha^2$ remains symmetric, since $(\alpha \alpha)^T = \alpha^T \alpha^T = \alpha \alpha$.

So it seems like the answer is yes. In fact I think $\alpha^n$ is positive definite for any $n \in \mathbb{Z}$. The even and odd $n$ are two different cases, using either the method you suggest or the one I use in the first paragraph.

You can also look at the eigenvalues, since the eigenvalues of $\alpha^2$ are just the squares of the eigenvalues of $\alpha$, and it's also diagonalizable.