# How to prove that $\langle U^HTUx , \ x\rangle = \langle Tx ,\ x\rangle$ for $\|x\| = 1$

Assume $U$ is a unitary matrix and $x\in\mathbb{C}^n$, $\|x\| = 1$
Prove: $\langle U^HTUx\ , \ x\rangle = \langle Tx\ ,\ x\rangle$

I know that $\langle U^HTUx\ , \ x\rangle = \langle TUx\ ,\ Ux\rangle$ ( the property of the inner product).

And I have a feeling that $(xU)^HT^HUx = x^HTx$ ( because $\|Ux\| = \|x\| = 1$, but I don't know how to prove it strictly.

Can you show me the derivation? Thanks!

---edit:

Is my statement wrong?

I am speaking Numerical range : $W(T) = \{\langle Tx,x\rangle :x\in\mathbb{C}^n;||x|| =1\} \subseteq\mathbb{C}$ and $W(U^HTU) = W(T)$

this is no true in fact in $\mathbb{R^2}$ let $$U=\left(\begin{matrix} i& 0\\ 0& -i \end{matrix}\right)$$ then $U$ is unitary matrix and $$T=\left(\begin{matrix} 1& 1\\ 1& 2 \end{matrix}\right)$$ we have : $$U^tTU=\left(\begin{matrix} -i& 0\\ 0& i \end{matrix}\right)\left(\begin{matrix} 1& 1\\ 1& 2 \end{matrix}\right)\left(\begin{matrix} i& 0\\ 0& -i \end{matrix}\right)=\left(\begin{matrix} -i& -i\\ i& 2i \end{matrix}\right)\left(\begin{matrix} i& 0\\ 0& -i \end{matrix}\right)=\left(\begin{matrix} 1& -1\\ -1& 2 \end{matrix}\right)$$ the conclusion come directly by enter link description here or for $x=(\frac{\sqrt 2}{2},\frac{\sqrt 2}{2})$ we have : $$\langle T x,x\rangle=(\sqrt2,\frac{3\sqrt 2}{2})\left(\begin{matrix} \frac{\sqrt 2}{2}\\ \frac{\sqrt 2}{2} \end{matrix}\right)=5/2$$ but $$\langle U^tTU x,x\rangle=(0,\frac{\sqrt 2}{2})\left(\begin{matrix} \frac{\sqrt 2}{2}\\ \frac{\sqrt 2}{2} \end{matrix}\right)=1/2$$

**Edit : ** For the second question this is true that numerical range is preserved under unitary maps, in fact this use this trivial lemma

$$\|x\|=1 \iff \|Ux\|=1$$

Proof : $$\|Ux\|^2=\langle Ux,Ux\rangle=\langle U^tUx,x\rangle=\langle x,x\rangle=\|x\|$$

so \begin{eqnarray} W(U^tTU)&=&\{ \langle U^tTUx,x\rangle ; \|x\|=1\}\\&=&\{ \langle U^tTUx,x\rangle ; \|Ux\|=1\}\\ &=& \{ \langle TUx,Ux\rangle ; \|Ux\|=1\}\\ &=&\{ \langle T\color{red}{y},\color{red}{y}\rangle ; \|\color{red}{y}\|=1\}\\ &=& W(T) \end{eqnarray}

• Hello! I have edited my question. Is there any mistake in my statement? But the last equation is true – stander Qiu Jul 29 '16 at 13:05

In $\mathbb{R}^2$, take:

• $x = (1, 0)^t$,
• $U$ = clockwise rotation by $90^\circ$,
• $T$ = matrix with $T_{11} = 1$ and zeros elsewhere.

Then $\langle T x, x \rangle \neq \langle U^H T U x, x \rangle$.

The numerical range of $T$ and $U^H T U$ is, of course, the same, because you can set $y = U x$ and loop over $y$ instead of $x$.

• Hello! I have edited my question. Is there any mistake in my statement? But the last equation is true – stander Qiu Jul 29 '16 at 13:05