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Suppose $A$ is an $n\times n$ complex matrix. How to show the following two properties

  1. If $\lambda$ is an eigenvalue of $A\bar{A}$, so is $\bar{\lambda}$. Here $\bar{A}$ means the entrywise conjugate of $A$.

  2. The algebraic multiplicity of negative eigenvalues (if any) of $A\bar{A}$ are even.

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For 1 I think it can be proved that the characteristic polynomial of $A\overline{A}$ has real coefficients. –  Beni Bogosel Apr 23 '12 at 19:53
    
I think this holds for any matrix $M$. I mean, suppose that $\lambda$ is an eingenvalue of $M$, then so is $\bar{\lambda}$. –  leo Apr 23 '12 at 19:58
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@leo: Only for matrices $M$ with real components. –  Beni Bogosel Apr 23 '12 at 19:59
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@leo: no, not for complex matrices in general. –  Robert Israel Apr 23 '12 at 19:59
    
Is the proof for the second property easy? –  Sunni Apr 23 '12 at 20:01
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3 Answers 3

up vote 5 down vote accepted
  1. It is known that the characteristic polynomial of $A\overline A$ is equal to the characteristic polynomial of $\overline AA$. Then $$ {\det(\overline AA-\lambda I)}=\det(A\overline A-\lambda I) $$ but for real numbers $\lambda$ the conjugate of the LHS is equal to the RHS, therefore the characteristic polynomial of $A\overline A$ has real coefficients.

  2. An idea would be to prove that the characteristic polynomial of $A\overline A$ takes positive values for $\lambda \geq 0$. Then, if there exists a negative eigenvalue $\mu<0$ with odd multiplicity, (we can consider $\mu$ to be the the greatest such eigenvalue) then for $\lambda\in (-\mu-\varepsilon,-\mu) \cap [0,\infty)$ (for some $\varepsilon$ small enough) the polynomial would take negative values and we have a contradiction. It remains to prove that $$ \det(A\overline A-\lambda I)\geq 0,\ \forall \lambda \geq 0 $$ One way to do this is to prove that all the coefficients of the characteristic polynomial are positive.

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what about the second one? –  Sunni Apr 23 '12 at 20:01
    
@Sunni, by negative eigenvalues do you mean real negatives eigenvalues? –  leo Apr 23 '12 at 20:11
    
Where is that known from, actually? –  Henning Makholm Apr 23 '12 at 20:15
    
@HenningMakholm: See for example: math.sc.edu/~howard/Classes/700/charAB.pdf I'm sure it has been posted on MS also. –  Beni Bogosel Apr 23 '12 at 20:21
    
Ah, of course. I somehow formed the impression that it was something specific to entrywise conjugation. –  Henning Makholm Apr 23 '12 at 20:26
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here's a try:

(edit, deleted my development for q1 which was wrong, my bad!, thanks to Henning Makholm)

  • If I'm not mistaken, you can show that the determinant is positive (because I think that you can show something like $\det(A\overline{A})=\det(A)det(\overline A)=\overline{\det A}^2$) which thus implies that the product of eigenvalue be positive. The product of $\lambda$ and $\overline{\lambda}$ is obviously positive if $\rm{Im}\lambda\neq 0$ but if you have a $\lambda\in\mathbb R$ and $\lambda<0$ then there must be another $\lambda'\in\mathbb R$ with $\lambda'<0$. This implies that there be an even number of real negative eigenvalues.

(edit, as Beni mentioned, this just shows that the total number of real negative eigenvalues is even (product is positive) which might not answer the question)

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Yes, but what you say doesn't prevent the existence of, let's say an eigenvalue -2 with multiplicity 1 and an eigenvalue -4 with multiplicity 3. You say that the product of ALL real eigenvalues is positive. But it is asked that the multiplicity of EVERY negative eigenvalue is positive. –  Beni Bogosel Apr 23 '12 at 20:18
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Um, how do you get $z_{ij}=|a_{ij}|^2$? It is not true for $A=\pmatrix{0&1\\1&0}$, for example. –  Henning Makholm Apr 23 '12 at 20:23
    
@HenningMakholm, yes absolutely, sorry, stupid mistake. thanks! –  tibL Apr 23 '12 at 20:26
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I noticed property 2 was contained in Theorem 1 in Djokovic, On Some Representations of Matrices, Linear and Multilinear Algebra, 1976, Vol. 4, pp. 33-40.

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