1
$\begingroup$

Let $A$ be an $n×n$ matrix over $\Bbb C$ (complex) and $F(A)$ the field of values. Let $U$ be an $n×n$ unitary matrix.

(i) Show that $F(U^*AU) = F(A)$.

I am not sure how to deal with equality of fields. How do we show that both fields are equal? Showing that if one element belong to one field, it belongs to the other field too?

$\endgroup$
3
  • 1
    $\begingroup$ What is the field of values of a matrix? $\endgroup$
    – Bernard
    Jan 3 at 18:24
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Numerical_range $\endgroup$ Jan 3 at 18:26
  • $\begingroup$ One has $U^*=U^{-1}$ because unitary. Note that for $n=1$ you do have $U^*AU=\dfrac 1aAa=A$ so the proof is trivial in this case. Try to go to $n\gt1$. $\endgroup$
    – Piquito
    Jan 3 at 18:53
0
$\begingroup$

First thing don't worry about field equality as this leads to unnecessarily abstract concepts. Both fields are subsets of complex numbers so you can treat this just as a question about equality of subsets. Sou you just prove two inclusions.

I think that for the complete proof you need these two observations:

1 Multiplication by unitary matrix does not change the norm of a vector see $(Ux)^*(Ux)=x^*U^*Ux=x^*Ix=x^*x$ where $I$ is the identity matrix.

2 Unitary matrix is regular. In other words, it is surjective as a linear operator on a vector space.

Now the equation $x^*U^*AUx=(Ux)^*A(Ux)$ will allow you to complete the proof. Just for each value in the one field of values find a vector that gives you this value and try to prove the existence of a vector that gives you the same value in the other field of values.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.