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Let $A$ and $B$ be $n\times n$ hermitian matrices. Then show that trace of $A^kB^k$ and trace $(AB)^k$ are real for any positive integer $k$.

Agu
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hafsah
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1 Answers1

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Hints.

$A^kB^k$ is the product of two Hermitian matrices $A^k$ and $B^k$.

$(AB)^k$ is the product of $A$ and another Hermitian matrix $B(AB)^{k-1}=BABAB\cdots ABAB$.

So, the essential problem is why $\operatorname{tr}(PQ)$ is real whenever $P,Q$ are Hermitian. Note that every Hermitian matrix is unitarily diagonalisable. Also, the diagonal entries of every Hermitian matrix (diagonal or not) are real.

user1551
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