unitarily similar characterization

Today at lesson, the professor brought up the notion of unitary similarity when talking about Schur theorem.

After lesson, we talked about complete characterizations of this similarity (not only for normal matrices, where it is easy). Turned out we didn't know any.

We know that two unitarily similar matrices must have

• same Jordan Form (since they must be similar)
• same singular values
• similar real and imaginary part (where the real part of a matrix $A$ is $A+A^*/2$ and the imaginary part $A-A^*/2\text{i}$)

Now I found written in a forum that the first two conditions are sufficient, but I can't find a proof or a counterexample, or a reference to this.

Actually the exact lemma cited was

$A,B$ are unitarily similar iff they are similar and $tr(A^*A)=tr(B^*B)$

Any help?

The statement is not true, in general. The forward implication holds for all complex matrices, but the reverse only holds for $2 \times 2$ matrices. This is dealt with in some detail in "Matrix Analysis" - Horn and Johnson in chapters 21 - 24.

$\Rightarrow$Forward implication

Suppose that $A$ and $B$ are unitarily similar, then there exists a unitary matrix $U$ such that, $$A = UBU^{*}$$ Using the cyclic property of the trace we have that, \begin{align} tr(A^*A) =& tr((UBU^*)^*(UBU^*)) \\ =& tr(UB^*U^*UBU^*) \\ =& tr(UB^*BU^*) \\ =& tr(U^*UB^*B) \\ =& tr(B^*B) \end{align} and so $tr(A^*A) = tr(B^*B)$, as required.

$\Leftarrow$ Reverse implication

Conversely, suppose that $tr(A^*A) = tr(B^*B)$, and that $A$ and $B$ are similar.

Here we can make use of Specht's theorem. Which states that two matrices $A$ and $B$ are unitarily equivalent if and only if $tr W(A,A^*) = tr W(B,B^*)$ for all words $W$.

It is clear that for $2 \times 2$ matrices $tr(A^*A) = tr(B^*B)$ is a sufficient condition, but for larger degrees we also need equality for higher order words, such as $tr(A^2A^*) = tr(B^2B^*)$.

The reverse implication is not true, in general. Consider the $3\times 3$ matrices, \begin{align} A = \left[\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \end{array}\right] & & B = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 5 \\ 0 & 0 & 3 \end{array} \right] \end{align}

$A$ is similar to $B$ through their eigenvalue decompositions, and $tr(A^*A) = tr(B^*B)$. But $tr(A^2A^*) \neq tr(B^2B^*)$, and so $A$ and $B$ are not unitarily equivalent.

• I knew that theorem, but it still doesn't answer my question. Commented Feb 26, 2018 at 2:08
• I have added some extra information. The book contains an exercise in which the reader is asked to find a counter example for $3 \times 3$ matrices. Commented Feb 26, 2018 at 7:56
• Thanks for the counter. Maybe I'll ask in another question, but if you could answer too would be great. In the original post I asked also "is it true that similarity and same singular values imply unitarily similarity?" Commented Feb 26, 2018 at 16:02
• Your counterexample doesn't sound right: $tr(A^\ast A)=\|A\|_F^2>\|B\|_F^2=tr(B^\ast B)$. Commented Feb 26, 2018 at 18:17
• @Exodd The matrices $A = \left[ \begin{array}{ccc} 1 & 3 & 0 \\ 0 & 2 & 4+i \\ 0 & 0 & 3-i \end{array} \right]$ and $B = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 4+i & 3-i \end{array} \right]$ are a counterexample. Commented Feb 27, 2018 at 8:04