Is the square root of positive definite endomorphism of an inner product space, also positive definite?

Question

Let $V$ be an inner product space. Let $\alpha$ be an endomorphism. $\alpha$ is positive definite, i.e. $\langle\alpha(x),x\rangle>0,\forall x\neq 0$, and $\alpha$ is self-adjoint. In my book, there is the following exercise

Show that $\alpha$ has a unique positive definite square root. [Note: whether the inner product space is over $\mathbb C$ or $\mathbb R$ is not specified.]

Then, I try to do an easier version of the problem, by letting $\beta^2 =\alpha$ and proving that $\langle \beta (x),x\rangle >0$. After some futile algebraic manipulation, I just get useless results like $$ \langle \beta(x),\beta^2 (x) +x\rangle=2\langle \beta (x),x\rangle, $$ which does not lead to anywhere.

I know that I can write out $\alpha$ in its matrix form (pick an orthogonal basis). In that way, I can easily compute the Jordan conical form, and calculate the square root by series expansion. But that doesn't seem to be the point of this question. I think I should find a more elegant solution.

Is there any method to solve this without using matrix?

Answer 1

A self-adjoint matrix can be diagonalized by a unitary transformation, i.e. there is a unitary $u$ such that $\Lambda = u \alpha u^*$ is diagonal, the diagonal elements being the eigenvalues of $\alpha$. $\alpha$ being positive definite, these diagonal elements are positive. The positive definite square root of $\alpha$ is then $u^* \Lambda^{1/2} u$, where $\Lambda^{1/2}$ is diagonal with diagonal elements the square roots of the diagonal elements of $\Lambda$.

Note, by the way, that it's not saying "$\alpha$ has exactly one square root, and that square root is positive definite". It's saying $\alpha$ has only one positive definite square root. There are also lots of square roots that are not positive definite.

The uniqueness of the positive definite square root can be proven using the polarization identity.