# A is normal if and only if every matrix unitarily equivalent to A is normal

I need to prove that if $A$ and $B$ are unitarily equivalent, then $A$ is normal if and only if $B$ is normal.

The proof is as follows:

Suppose $A$ is normal and $B = U^*AU$, where $U$ is unitary. Then $B^∗B = U^∗A^∗AU = U^∗AA^∗U = U^∗AUU^∗A^∗U = BB^∗$. If $U^∗AU$ is normal then it is easy to see that $U^∗AA^∗U = U^∗A^∗AU$. Multiply this equation on the right by $U^∗$ and on the left by $U$ to obtain $AA^∗ = A^∗A$.

However, I don't see how $B^*B= U^∗A^∗AU$.

$B^*B=(U^*AU)^*(U^*AU)$, and because $U$ is unitary $U^{-1}=U^*$ but I don't know how to continue. What am I missing?

• What you are missing is $(XY)^*=Y^*X^*$. Commented Nov 27, 2014 at 11:48
• Oh, I see now. Yeah the transpose will switch around the matrices. Thanks! Commented Nov 28, 2014 at 11:05

Let $A$, $B$ given with $B=U^*AU$ with $UU^*=U^*U=I$. Then $B^*=(U^*AU)^*=U^*A^*U$ and $$BB^*=(U^*AU)(U^*A^*U)=U^*AA^*U, \quad B^*B =U^*A^*UU^*AU = U^*A^*AU .$$ Let $A$ be normal. Then by the above calculations $$BB^*=U^*AA^*U=U^*A^*AU=B^*B.$$ If $B$ is normal then $$AA^* = UBB^*U^*= UB^*BU^* = A^*A.$$