# Show that $T$ is unitary if and only if we have an orthonormal basis

Let $\{v_1,...,v_n\} \subset V$ be an orthonormal basis of $V$. Show that the set $\{T(v_1),\dots,T(v_n)\}$ is an orthonormal basis of $V$ if and only if $T$ is unitary.

I have probably gone into more detail than is really necessary but I'm trying to get my head around all of these ideas.

Suppose $T$ is unitary. So we have that $\langle Tv,Tw \rangle = \langle v,w \rangle$ for any $v,w \in V.$

So $\langle Tv_i,Tv_j \rangle=\langle v_i,v_j \rangle=\delta_{ij}$. Thus we have $n$ orthogonal vectors in a space of dimension $n$ and so we have an orthonormal basis.

Now suppose we have an orthonormal basis. So $\langle Tv_i,Tv_j \rangle=\delta_{ij}.$

Let $v=\sum_{i=1}^n \langle v_i,v \rangle v_i$ and let $w=\sum_{j=1}^m \langle v_j,w \rangle v_j.$

$$\langle Tv,Tw \rangle=\langle \sum_{i=1}^n \langle v_i,v \rangle Tv_i, \sum_{j=1}^m \langle v_j,w \rangle Tv_j \rangle = \sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle \langle Tv_i,Tv_j \rangle=\sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle \delta_{ij}=\sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle =\langle v,w \rangle.$$ Thus $T$ is unitary.

Could someone verify if this is correct or not please?

Definition of $v$:

$$v=\sum_{i=1}^n \lambda_iv_i$$

Then consider the inner product with $v_i$:

$$\langle v_i,v \rangle =\langle v_i,\sum_{i=1}^n \lambda_iv_i \rangle=\sum_{i=1}^n\lambda_i \langle v_i,v \rangle = \sum_{i=1}^n \lambda_i \delta_{ij}=\lambda_i$$

• I'm a bit confused about how you define $v$ and $w$. Isn't that a recursive definition? Why is $v$ in the definition of $v$? – Noble Mushtak Apr 23 '16 at 11:38
• @NobleMushtak I've made an edit discussing where my definition came from. – Si.0788 Apr 23 '16 at 11:46
• OK, that makes more sense. Thanks! – Noble Mushtak Apr 23 '16 at 11:46
• @NobleMushtak Does it look fine apart from that? – Si.0788 Apr 23 '16 at 11:52
• I think this is basically correct, but why in the calculations in the central part of your post you took the conjugate of the first variable, not of the second one? – DonAntonio Apr 23 '16 at 11:53

## 1 Answer

This looks correct to me except for one thing:

$$\sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle \delta_{ij}=\sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle =\langle v,w \rangle.$$

At this last part, all of the terms where $i \neq j$ are cancelled out by $\delta_{ij}=0$. Thus., the second summation should just be from $i=1$ to $n$, so this is correct:

$$\sum_{i,j} \overline {\langle v_i,v \rangle} \langle v_j,w \rangle \delta_{ij}=\sum_{i=1}^n \overline {\langle v_i,v \rangle} \langle v_i,w \rangle =\langle v,w \rangle.$$

Other than that, I think this is a really good proof!