# Prove that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is an orthonormal basis for $V$

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 : $TT^*=T^*T=I$.

• This result is clearly false for a general linear operator $T$. Are you perhaps assuming $T$ is unitary (or an orthogonal map in the real case)? – Ted Shifrin May 29 '13 at 14:28
• The claim is wrong, for example if $T$ is not injective or not orthogonal – Hagen von Eitzen May 29 '13 at 14:29
• Given your condition that $<T(v_i), T(v_j)> = <v_i, v_j>$, what properties of $T$ are there? Do you want $T$ to be a unitary operator? – Calvin Lin May 29 '13 at 14:34
• The claim is precisely the characterization of orthogonal maps (which, of course, are bijective) so it can't be true in general. – DonAntonio May 29 '13 at 14:35
• @DonAntonio You're right. Then why $T(\beta)$ can be a basis in here? – noname May 29 '13 at 14:44

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 $V$.
• In fact that statement gives way more, @Adolfo : $\,T\,$ the is an orthogonal map (unitary in the complex case). – DonAntonio May 29 '13 at 14:37