Schur's Lemma says, any square matrix $A\in M_n $ can be decomposed to the following form:
$$ A=UTU^* $$
where $U$ is unitary and $T$ is upper triangular with $A's$ eigenvalues on its diagonal.
now if two matrices $A,B$ are similar (that is, they share the same eigenvalues), can one say that their Schur's decomposition results a mutual upper triangular matrix $T$?
$$ (\star)\; A\sim B \;\overset{?}{\Rightarrow}\; U^*AU=V^*BV $$
namely, does $(\star)$ hold?
My guess is that it's true by decomposing $A,B$ with schur's to the following forms:
$$ A =UT_1U^* \,,\, B=VT_2V^* $$
where $T_1$ and $T_2$ has the same main diagonal (same eigen values), but I can't continue proving that they are equal.
Is it somewhat related to the Jordan canonical form? are those T's the Jordan canonical forms of the matrices? I couldn't find any relationships between Shur's Lemma and Jordan's canonical form but I have a feeling its related and that is the continuation of the proof (if its even true)
thanks ahead!
Best Answer
The answer is negative. If $UAU^\ast=VBV^\ast$ for some unitary matrices $U$ and $V$, then $A$ and $B$ have the same Frobenius norm (defined by $\|X\|_F=\sqrt{\operatorname{tr}(X^\ast X)}$, i.e. it is the Euclidean norm of the vector $\operatorname{vec}(X)$ obtained by flattening the matrix $X$).
However, similar matrices don't necessarily have identical Frobenius norms. E.g. the Jordan block $A=\pmatrix{0&1\\ 0&0}$ is similar to $2A$, but $\|A\|_F=1\ne2=\|2A\|_F$.