Let $U$ be a unitary matrix, show that $r(x,y) := x^*Uy$ is an inner product.

Question

Let $U$ be a unitary matrix, show that $r(x,y) := x^*Uy$ is an inner product satisfying

1. $(u,v) = \overline{(v,u)}$

2. $(u,u)> 0$ for $u\neq0$; $(u,u)=0$ for $u= 0$

3. $(u+sv,w)=(u,w)+s(v,w)$

for a complex vector space $V$

Explain why this would not work if $U$ is simply invertible

Note: A matrix $U$ is unitary if $U^*U = I$

Don't know how to start this question... Hope someone can help. Thank you very much!

Answer 1

This should not work, matrices to generate scalar products are self-adjoint positive definite (SPD).

Indeed, property 1) translates to $U^*=U$, property 2) to $U>0$ and 3) is generally true for this type of construction.

Answer 2

Let's examine your claim, where I guess you define $x^*$ as the conjugate transpose. Now $$ r(v,u)=v^*Uu=\overline{u^*U^*v}=\overline{r(u,v)} $$ if and only if $$ u^*U^*v=u^*Uv $$ for all $u$ and $v$. This is the same as requiring that $U^*=U$, so $U$ must be Hermitian, not unitary.

However a Hermitian unitary matrix is not really interesting: the eigenvalues of a unitary matrix have modulus $1$, those of a Hermitian matrix are real. Thus a Hermitian unitary matrix can only have $1$ and $-1$ as eigenvalues. If you require it to define an inner product, the eigenvalues must be positive, so you just get the identity.