# Prove this field equality.

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?

• What is the field of values of a matrix? Jan 3 at 18:24
• en.wikipedia.org/wiki/Numerical_range Jan 3 at 18:26
• 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 $n\gt1$. Jan 3 at 18:53

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.
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.