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?