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Suppose a square matrix $A$ has eigenvalues $\lambda_1,...,\lambda_n$.

Note that A is a $n\times n$ matrix where $n$ is even.

Let $\widehat{A}$ be a matrix that is obtained from $A$ only by multiplying every second row in $A$ by $-1$. Is there a relationship between the eigenvalues of $A$ and $\widehat{A}$?

I don't think there is, but I am not sure.

eggSand
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    Some people use the notation $A'$ to mean the conjugate-transpose of $A$ (so that if all entries are real it's just the transpose). So a different notation might be more felicitous here. $\qquad$ – Michael Hardy Mar 29 '16 at 18:07
  • Is that better? Or does it also mean something else? – eggSand Mar 29 '16 at 18:09
  • $\ldots,$or maybe $\widetilde A$? $\qquad$ – Michael Hardy Mar 29 '16 at 18:14
  • Is it $b$ with the same row number as $A$ multiplied by $-1$? – eggSand Mar 29 '16 at 18:16
  • Only every second row of $A$ by $-1$ – eggSand Mar 29 '16 at 18:17
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    So $\widehat{A} = AD$ where $D$ is a diagonal matrix? – Squid Mar 29 '16 at 18:17
  • @Squid no $\hat{A}$ is obtained by multiplying every second row of $A$ by $-1$. – eggSand Mar 29 '16 at 18:18
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    Expanding on squids's comment: we have $\hat{A} = DA$ with $D = \text{diagonal}(1,-1,1,-1,\ldots)$. – Winther Mar 29 '16 at 18:22
  • I'm not sure if it matters, but is $n$ even or odd? – Paul Mar 29 '16 at 18:25
  • $n$ is always even here – eggSand Mar 29 '16 at 18:25
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    Take $A=\left( \begin{array}{cc} 1 & 1 \ -1 & 1 \ \end{array} \right)$ then $\hat{A} = \left( \begin{array}{cc} 1 & 1 \ 1 & -1 \ \end{array} \right)$. The eigenvalues are $\lambda_A = 1\pm i$ and $\lambda_{\hat{A}} = \pm \sqrt{2}$ so the eigenvalues are not the same in general. However we have some relationships. For example the product of all eigenvalues of $A$ will be (up to a sign) the same as the product of the eigenvalyes of $\hat{A}$ (the sign is the same if the matrix dimension $n$ is divisible by $4$ and $-1$ otherwise). – Winther Mar 29 '16 at 18:28
  • The original eigenvalue problem is defined by – Vini Mar 05 '17 at 21:09

1 Answers1

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Write $D=\text{diag}(1,-1,1,-1,\dots)$, so that $\hat{A} = DA$. Then we have $$\det(\hat{A}) = \det(D)\det(A) = (-1)^{n/2}\det(A),$$ so the product of the eigenvalues of $A$ is the same as that of $\hat{A}$, up to a sign.

In general, you can't expect strong relationships among the eigenvalues of $A$ and $\hat{A}$, for the following reason: Choose any two sets of eigenvalues $\lambda_1,\dots,\lambda_n$ and $\chi_1,\dots,\chi_n$. Naively we might expect it should be possible to find a matrix $A$ s.t. $A$ has the first set of eigenvalues and $\hat{A}$ has the second. This is because $\lambda$ is an eigenvalue of $A$ iff $\det(A-\lambda) = 0$, so we are effectively solving $2n$ equations (one for each of $\lambda_i$ and $\chi_i$) in the $n^2$ coefficients of $A$. Provided $n\geq 2$ (so that $n^2\geq 2n$) , we expect a solution to exist.

This argument is not rigorous, because the equations involving $A$ and those involving $\hat{A}$ need not be independent. (Indeed, we know it has to be false because of the determinant relationship above.) However, some experiments in Wolfram Alpha suggest that e.g. for $n=2$ it is possible to choose all but one of the eigenvalues of $A$ and $\hat{A}$ independently. Perhaps the determinant relationship is the only one that exists among the eigenvalues of $A$ and $\hat{A}$?

However, for some special classes of matrices we can find further relationships: Note that $\hat{A}^\dagger\hat{A} = A^\dagger D^\dagger D A = A^\dagger A$. For a normal matrix $M$, if $M$ has eigenvalues $\lambda_i$ then $M^\dagger M$ has eigenvalues $|\lambda_i|^2$. This means that if $A$ and $\hat{A}$ are both normal, then their eigenvalues must be equal up to complex phases.

Yly
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