This the third exercise in section 7.D from Axler's book (Linear Algebra Done Right).
Suppose $T \in \mathcal{L}(V)$. Prove that there exists an isometry $S \in \mathcal{L}(V)$ such that
$$T = \sqrt{TT^*}S.$$
$V$ is a finite-dimension inner product space.
I'm don't have any ideas on how to begin this. Any tips?
Well, it was considerably easier than I thought.
By the Polar Decomposition, there exists an isometry $S \in \mathcal{L}(V)$ such that $T^* = S\sqrt{TT^*}$. Taking the adjoint of each side, we get
$$ T = (S\sqrt{TT^*})^* = (\sqrt{TT^*})^*S^* = \sqrt{TT^*}S^*, $$
where the last equality follows because $\sqrt{TT^*}$ is self-adjoint. This yields the desired result, because $S^*$ is also an isometry.