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Problem: Let $A_{n\times n}$ and $B_{n\times n}$ be complex unitary matrices, where n is an odd number. Prove that the number: $$z=\det(A+B) \det(\overline A-\overline B)$$ is purely imaginary.

My idea:

We have that $AA^*=I$ and $BB^*=I$.

We also have that $\det A=\det A^*=(-1)^n \det A^*=-\det A$. Same for matrix $B$.

Am I free to say $A$ and $B$ are skew-symmetric matrices?

$$z=\det(A+B)\overline {\det(A-B)}$$ $$z=\det((A+B) {(A-B)^*)}$$ $$z=\det((A+B) {(A-B)^*)}$$ $$z=\det((A+B) {(A^*-B^*))}$$ $$z=\det((AA^*-AB^*+BA^*-BB^*)$$ $$z=\det((I-AB^*+BA^*-I)$$ $$z=\det((BA^*-AB^*)$$

This is where I am unsure of how to proceed. Am I free to say since matrices are unitary and skew-symmetric, their eigenvalues are purely imaginary thus $z$ must be as well? Is this statement correct? Thank you all in advance.

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2 Answers 2

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Hint: $(BA^*-AB^*)^*=AB^*-BA^*=-(BA^*-AB^*)$ thus its eigenvalues are purely imaginary. An the product of an odd number of purely imaginary is purely imaginary.

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    $\begingroup$ To make an argument that doesn't depend on eigenvalues: note that $$ \overline{\det(BA^* - AB^*)} = \\ \det([BA^* - AB^*]^*) = \\ \det((-1)(BA^* - AB^*)) =\\ (-1)^n \det(BA^* - AB^*) $$ $\endgroup$ Jun 22, 2016 at 19:49
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As proceeding along the lines given in the hint by @Tsermo , we find that $$ z = det(BA^{*} -AB^{*}) =\Pi_i^{n} e_i $$ where $e_i$ are the eigenvalues of $(BA^{*}-AB^{*}) $ which are purely imaginary(as the matrix$(BA^{*}-AB^{*}) $ is skew symmetric of odd order).Since n is odd therefore the result follows. By the way how do you say that A and B are skew symmetric? In fact $det (A) = \frac{1}{det(A^{*})}$

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  • $\begingroup$ Yes, I understand that. I was unsure about the result so I combined some silly ideas from previous problems where similar conclusion was made when told $n$ is an odd number. Thank you. $\endgroup$
    – Asleen
    Jun 22, 2016 at 19:31

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