# $U$ invertible and $U^{\ast}U=1$ then $UU^{\ast}=1$ (Barry Simon)

(a) Let $$U$$ map $$H\to K$$ for two Hilbert spaces. Prove that $$|U\varphi|=|\varphi|$$ for all $$\varphi$$ if and only if $$U^{\ast}U=1$$

(b) If $$U$$ is invertible and $$U^{\ast}U=1$$, prove that $$UU^{\ast}=1$$ (Hint: Multiply by $$U$$ and $$U^{-1})$$

For (a) i have this:

$$|U\varphi|^2=|\varphi|^2\Leftrightarrow \langle U\varphi,U\varphi\rangle=\langle \varphi,\varphi\rangle \Leftrightarrow \langle\varphi, U^{\ast}U\varphi\rangle =\langle\varphi,\varphi\rangle \Leftrightarrow U^{\ast}U\varphi=\varphi \Leftrightarrow U^{\ast}U=1$$

For (b) how prove (b)?

• As the hint says, multiply by $U$ and $U^{-1}$. – anomaly Oct 21 '18 at 1:13
• oh! $U^{\ast}U=1$ then $UU^{\ast}U=U$ then $UU^{\ast}UU^{-1}=UU^{-1}=1$ then $UU^{\ast}=1$. If $U$ invertible and $U^{\ast}U=1$ then $U^{-1}=U^{\ast}$? – eraldcoil Oct 21 '18 at 1:24

Because you don't mention it, you might be overlooking something in your argument for $$(a)$$. You don't say how you get from $$\tag1\langle \varphi, U^*U\varphi\rangle=\langle \varphi,\varphi\rangle$$ to $$U^*U\varphi=\varphi.$$ This is not hard, but it is not entirely immediate. You would usually use the polarization identity to get from $$\langle \varphi,(U^*U-I)\varphi\rangle=0$$ (for every $$\varphi$$) to $$U^*U=I$$. And the argument does not work without quantifiers: you cannot conclude, if you have $$(1)$$ for a single $$\varphi$$, that $$U^*U\varphi=\varphi$$. You need to have $$(1)$$ for all $$\varphi\in H$$.
For instance, with $$H=K=\mathbb C^2$$, let $$\varphi=\begin{bmatrix} 1\\0\end{bmatrix}$$, and $$T=\begin{bmatrix} 1&0\\-1&0\end{bmatrix}$$. Then $$\langle T\varphi,\varphi\rangle=\left\langle\begin{bmatrix} 1\\-1\end{bmatrix},\begin{bmatrix} 1\\0\end{bmatrix}\right\rangle=1=\langle\varphi,\varphi\rangle,$$ but $$T\varphi\ne\varphi$$.
For part (b), if $$U^*U=I$$ and $$U$$ is invertible, then multiplying on the right by $$U^{-1}$$ you get $$U^*=U^{-1}$$. Now you have $$I=UU^{-1}=UU^*$$.