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Let $A$ is a normal matrix: $A^*\! A = A A^*\!\!$,$\,$ and $AB = BA$. Prove that $A^*\!B=BA^*\!\!$.

I can prove that if $\det A\ne 0$ by multiplication $AB=BA$ by $A^*$ left and right and using some manipulation. But I have no idea what to do if $\det A = 0$.

Algebraic Pavel
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Michael Galuza
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3 Answers3

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Hint: If $A$ is normal then $A^*$ is a polynomial in $A$.

$\bf{Added:}$

Since $A$ is normal there exists $U$ a unitary matrix so that $A = U D U^*$ ( $A$ is unitarily diagonalizable), and so $A^* = U \bar D U^*$. Let $\lambda_k$ be the eigenvalues of $A$ ( the diagonal of $D$). Take $P$ a polynomial with complex coefficients so that $$P(\lambda_k) = \bar \lambda_k$$ Check that $P(A) = A^*$

orangeskid
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We can use the fact that $$\def\tr{\mathrm{tr}} X=0\iff \tr(XX^*)=0. $$ Since $A$ and $B$ commute, $A^*$ and $B^*$ commute as well. Together with the cyclic property of trace, $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, we find that in each term of $$ \begin{split} \tr[(A^*B-BA^*)(A^*B-BA^*)^*] &= \tr(A^*BB^*A)+\tr(BA^*AB^*) -\tr(BA^*B^*A)-\tr(A^*BAB^*) \end{split} $$ is equal to a constant, say, $\tr(A^*AB^*B)$. E.g., $\tr(A^*BB^*A)=\tr(AA^*BB^*)=\tr(A^*AB^*B)$ and $\tr(BA^*B^*A)=\tr(BB^*A^*A)=\tr(A^*ABB^*)=\tr(A^*AB^*B)$. Hence the trace of $(A^*B-BA^*)(A^*B-BA^*)^*$ is zero.

Algebraic Pavel
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In fact, if $|A|\neq 0$, then we can find a sequence $A_n$ such that $\lim_{n\rightarrow \infty}A_n=A$ and every term $A_n$ is inverse, that is, $|A_n|\neq 0$.

SHY
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  • and what? How it helps me? – Michael Galuza Jun 04 '15 at 05:46
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    @MichaelGaluza I guess that it was supposed to have $|A|=0$ instead of $|A|\neq 0$. Since the subset of nonsingular matrices is dense, any singular matrix can be written as a limit of a sequence of nonsingular ones. Since $f(X):=X^B-BX^$ is continuous in $X$, if $A_n\to A$ as $n\to\infty$ then $f(A)=\lim\limits_{n\to\infty}f(A_n)=0$. – Algebraic Pavel Jun 04 '15 at 14:10