I'm trying to prove the next:
Let $B$ be a nonsingular $n$-rowed real square matrix and $C = BB^{T}.$ Then $C$ has a nonsingular positive square root A.
I'm confused about it. I'd like to apply the same theorem for positive, selfadjoint, linear, bounded operators to ensure the existence of such positive square root. So I have to prove those properties but I don't know how to do this. A way that I've thought is considering the linear map associated to matrix $C$ and work with it, but I think this is not necessary because it's a lot of work, I guess there is another way easier to prove it.
For example, $C$ is selfadjoint because the adjoint on matrix is transpose matrix, but to check is a "positive operator", I don't know how to prove it because of the inner product. Even this, if we have such properties, how can we prove positiv eswuare root is nonsingular?
Any kind of help is thanked in advanced.
Hint: You can compute $$ \langle Cx, x \rangle = (Cx)^t x\\ = ((BB^t)x)^t x \\ = x^t (BB^t)^t x \\ = x^t (BB^t) x \\ = (x^t B)(B^t x) \\ \langle B^tx, B^tx \rangle $$