# 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 $v$? Apr 23, 2016 at 11:38
• @NobleMushtak I've made an edit discussing where my definition came from.
– MHW
Apr 23, 2016 at 11:46
• OK, that makes more sense. Thanks! Apr 23, 2016 at 11:46
• @NobleMushtak Does it look fine apart from that?
– MHW
Apr 23, 2016 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? Apr 23, 2016 at 11:53

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