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

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?

• Your matrix is not positive definite. Nov 29, 2019 at 0:35
• Ok. I have realised that as well. Nov 29, 2019 at 0:37

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.