Let $S$ be an isometry with adjoint $S^\ast$. Prove that $S^\ast S=SS^\ast=I$.
Here is what I have so far:
Since $S$ is isometry, there is an orthonormal basis of $e_j$'s such that $\|Se_j\|=\|e_j\|$.
Then $\langle e_j,e_k\rangle=\langle Se_j,Se_k\rangle=\langle e_j,S^\ast Se_k\rangle$ which implies that $S^\ast Se_j=e_j$ whch means $S^\ast S=I$.
How do I prove that $SS^\ast =I$ as well though?