Let $T$ be a linear operator on a finite diemnsional inner product space $V$.
Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$.
Proof
Let $\beta=\{v_1,\dots,v_n\}$ be an orthonormal basis for $V$ so $T(\beta)$.
$\langle T(v_i),T(v_j)\rangle=\langle v_i,v_j\rangle =\delta_{ij}$ , therefore $T(\beta)$ is an orthonormal basis for $V$.
The first equality is from the equivalent theorem $\langle T(x),T(y)\rangle=\langle x,y\rangle$.
I wonder why the last sentence means that $T(\beta)$ is an orthonormal basis for $V$.
They are linearly independent, of course, but they do span $V$? How can show that?
Sorry to make you be confused. I omitted this information : $TT^*=T^*T=I$.
Best Answer
So, now that you've added the hypothesis that $T$ is unitary, it makes sense. You've shown $T$ takes an orthonormal set of vectors to another orthonormal set. If $\{v_j\}$ is a basis, then $\dim V=n$, and so the $n$ linearly independent vectors $T(v_j)$ must likewise be a basis, as they span an $n$-dimensional subspace of $V$.