For $U$ unitary and $U+iI$ hermetic, prove: $U = -iI$

Question

Let $U$ be a unitary matrix $nxn$ such that $U+iI$ is hermetian.

Prove:

$$ U = -iI $$

What i tried (And got stuck...)

If i show that $\forall u \in V$:

$$ <(U+iI)u,u> = 0 $$

Because (U+iI) is hermetic, i will be able to conclude:

$$ (U+iI) = 0 \Rightarrow U = -iI $$

As needed.

So:

$$ <(U+iI)u,u> = \\ <Uu+iu,u> = \\ <Uu,u> + i<u,u> = \\ <Uu,u>+i||u||^2 = \\<Uu,u>+i||Tu||^2 = \\ <Uu,u> + i<Uu,Uu> $$

But i dont see how can i get to zero here...

Another way that i tried is using hermetic rules:

$$ \forall u,v \in V <(U+iI)u,u> = <u,(U+iI)u> $$

I got to: $$ 2i<v,u> = <v,Uu> - <Uv,u> $$

I tried to choose $v = u$, but still i dont get too much...

Another try was to look at the norm $$ ||(U+iI)|| = <U+iI,U+iI> $$ I whished it will turn out zero, but i got: $$ <U+iI,U+iI> = <U,U> + <I,I> $$ Now what?

Those are my homework, so better a hint than a solution.

Thanks for all the answers.

Answer 1

Do you know the Spectral Theorem? If so, since the matrix $A:=U+iI$ is Hermitian, you know that there exist a diagonal matrix $D=diag(\lambda_1,...,\lambda_n)$ consisting of the eigenvalues for $A$ and an invertible $P$ such that $A=PDP^{-1}$. Now try to use/show that

(i) Eigenvalues of a Hermitian matrix are real and (ii) Eigenvalues of unitary matrices have modulus 1,

to conclude that all the $\lambda_i$'s are zero.

Answer 2

Hint: Show that the spectral radius of $U+iI=0$. Since $U+iI$ is Hermitian this is enough prove the result. You will need the facts that spectrum of a unitary operator is contained in the unit circle and the spectrum of a Hermitian operator is contained in the real line.

Note: the spectrum of a matrix is nothing but the set of eigen values and spectral radius is the maximum value of$|\lambda|$ over all eigen values $\lambda$.

Answer 3

Hint: show that any element $A$ can be uniquely written as $X+iY$ with $X,Y$ Hermitian (or hermetic, however you like to call them, as long as this means $X=X^*, Y=Y^*$) (take $X=(A+A^*)/2, Y=(A-A^*)/2i)$

Write $U=(U+iI)+i(-I)$. Conclude that $U+U^*=2(U+iI)$ and therefore $U^*=U+2iI$. Multiplying by $U$, we have $U^2+2iU-I=0$. Thus $(U+iI)^2=0$.

Can you see why $U+iI=0$ from here?