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If two square matrices, $A$ and $B$, of the same size and with complex entries commute, why does $\overline{A}$ (the complex conjugate of $A$) commute with $B$? If this is not true, is it true if one is nilpotent?

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  • $\begingroup$ This doesn't answer the question, but $\bar{A}$ commutes with $\bar{B}$ since $0=\bar{0}=\overline{AB-BA}=\bar{A}\bar{B}-\bar{B}\bar{A}$. $\endgroup$ – vadim123 Feb 17 '14 at 2:43
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If $A$ commutes with $B$, then $\overline{A}$ commutes with $\overline{B}$. If $B$ is real, then $\overline{B} = B$ and your claim is true. However, this is not true in general. For example, $A$ commutes with itself, but $A$ does not necessarily commute with $\overline{A}$. For a nilpotent example, take $$A = \pmatrix{0&i&0\\0&0&1\\0&0&0}$$ and check that $A \overline{A} \ne \overline{A} A$.

What's going on: If $A$ commutes with $B$, and $v$ is an eigenvector of $B$ with eigenvalue $c$, then $B(Av) = A(Bv) = c(Av)$. Thus $Av$, if it is nonzero, is another eigenvector of $B$ with the same eigenvalue $c$. Now if you replace $A$ with $\overline{A}$, there is no reason to expect that $\overline{A} v$ will be an eigenvector of $B$ with eigenvalue $c$, just because $Av$ is.

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