# Math Help - Proofs, linear transformation is unitary

1. ## Proofs, linear transformation is unitary

Let $T: V \to V$ be the linear transformation given by $T(\vec x)=c \vec x$ where $c\in \mathbb{C}$. Demonstrate that T is unitary if and only if $|c|=1$.
If $V$ is a unidimensional vector space, demonstrate that the only unitary linear transformations are the one given previously. In particular, if $V$ is a unidimensional vector space, there exist only 2 orthogonal transformations: $T(x)=x$ and $T(x)=-x$.
Attempt: Not much. I want to prove $\Rightarrow$ first.
If T is unitary, then $T^{-1}(\vec x)=T^{\dagger}(\vec x)$. I don't know how to go further, I must show that this implies that $|c|=1$.
Hmm... $T^{-1} (\vec x)=\frac{\vec x}{c}=x^{\dagger} c^{\dagger}$.
Stuck here.

2. You know that unitary transformations preserve length, don't you? Can't you use that?

3. Originally Posted by Ackbeet
You know that unitary transformations preserve length, don't you? Can't you use that?
Well I wasn't aware of that. My freshman course in L.A. wasn't made for physicists so we didn't cover unitary transformations in particular.
So it means that $|\vec x|=|c \vec x|=|c||\vec x|$. Assuming $\vec x \neq 0$, we get $|c|=1$ as required.
For $\Leftarrow )$ I don't really know where to start or better said: what property to use.

4. If you can use "a linear transformation is unitary if and only if it preserves lengths (||Ax||= ||x||)" then look at ||Ax||= |c|||x|.

If not the what is your definition of unitary?

5. Originally Posted by HallsofIvy
If you can use "a linear transformation is unitary if and only if it preserves lengths (||Ax||= ||x||)" then look at ||Ax||= |c|||x|.

If not the what is your definition of unitary?
I think the definition we're using is "T is unitary if and only if its matrix representative is unitary". Our definition of a unitary matrix is $A^{-1}=A^{\dagger}$.
But I'll try to follow your tip anyway. If I have any problem I'll repost.
Edit: Ok so if I understand well, A would be the matrix representative of T. We have ||Ax||=|c||x|=|x|, therefore T preserves lengths and so is unitary. Is this a valid proof?

6. Ok, so suppose matrix $A$ is unitary. That is, assume $A^{-1}=A^{\dagger}.$ Let $x$ be a nonzero vector. Then

$\|Ax\|^{2}=\langle Ax|Ax\rangle=\langle x|A^{\dagger}Ax\rangle
=\langle x|A^{-1}Ax\rangle=\langle x|Ix\rangle=\langle x|x\rangle=\|x\|^{2}.$

Therefore, since $\|Ax\|\ge 0$ and $\|x\|>0,$ we may conclude that $\|Ax\|=\|x\|$ for all nonzero $x$. If $x=0,$ the result is also true.

This proof you can use as an intermediate step in the proof you're ultimately interested in.

Does this help?

7. Yes it helped Ackbeet. If I'm not wrong, you're reasoning is for the $\Rightarrow )$ part of the proof, right? What you've showed is equivalent to say $\|Ax\|= \|cx\|= |c| \|x\|=\|x\| \Rightarrow |c|=1$, for all x. Which completes the proof for the $\Rightarrow )$ part. I hope I'm not wrong on this.
For part $\Leftarrow )$, is my "proof" of post #5 valid?

8. Right. I think you've correctly finished the forward direction, if you include the lemma I proved in your proof.

The reverse direction I'm not so sure of. You have to show that, assuming there exists $c\in\mathbb{C}$ such that $|c|=1$, such that $T\vec{x}=c\vec{x}$ for all vectors $\vec{x}$, that $T^{-1}=T^{\dagger}.$

So, assume the hypotheses with all the labels the same. I think you might also be able to assume that $T^{-1}$ exists. You must show that $T^{\dagger}T=TT^{\dagger}=I.$

Can you compute $T^{-1}$ and $T^{\dagger}$ directly? I think you might be able to do that. Then you simply show that they are the same.