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Let $A \in \mathbb{R}^{n\times n}$ and $\mu_1, \mu_2, \dots, \mu_n$ be the eigenvalues of $A^TA$. Let

$$B := \left[\begin{array}{cc} 0 & A^T\\ A & 0\end{array}\right]$$

What is the relationship between the eigenvalues of $A^TA$ and $B$?

Using SVD decomposition, I can solve this question. I want to know whether there is another solution.

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  • $\begingroup$ If $A=U \sum V^T$ is its SVD. Let $\sum=\text{diag}(\mu_1,\mu_2,\ldots,\mu_n)$, $U=[u_1,\ldots,u_n]$ and $V=[v_1,\ldots,v_n]$. Then eigenvalues of $B$ are $\pm\mu_i$ with corresponding unit eigenvectors $\frac{1}{\sqrt{2}}\left[\begin{array}{c} v_i\\ \pm u_i \end{array}\right]$. $\endgroup$
    – Math-Data
    Nov 15, 2016 at 10:40
  • $\begingroup$ Sorry, I have edited the title. $\endgroup$
    – Math-Data
    Nov 15, 2016 at 10:42
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    $\begingroup$ That is wrong. The $\mu$'s are the eigenvalues of $\rm A^{\top} A$. $\endgroup$ Nov 15, 2016 at 10:49
  • $\begingroup$ Yes, you are right. If $\sum=\text{diag}(\sqrt{\mu_1},\sqrt{\mu_2},\ldots,\sqrt{\mu_n})$. Then eigenvalues of $B$ are $\pm\sqrt{\mu_i}$. $\endgroup$
    – Math-Data
    Nov 15, 2016 at 10:56

2 Answers 2

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Since $\mathrm A^{\top} \mathrm A$ is positive semidefinite, its eigenvalues, $\mu_1, \dots, \mu_n$, are nonnegative.

The characteristic polynomial of $\mathrm B$ is

$$\det (s \mathrm I_{2n} - \mathrm B) = \det \begin{bmatrix} s \mathrm I_{n} & -\mathrm A^{\top}\\ -\mathrm A & s \mathrm I_{n} \end{bmatrix} = \det (s^2 \mathrm I_{n} - \mathrm A^{\top} \mathrm A)$$

Thus, the $2n$ eigenvalues of $\mathrm B$ are $\pm\sqrt\mu_1, \dots, \pm\sqrt\mu_n$.

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  • $\begingroup$ Yes, Thank you so much. This is completely right. How do you obtain eigenvectors? $\endgroup$
    – Math-Data
    Nov 15, 2016 at 11:05
  • $\begingroup$ If you also want the eigenvectors, better use the SVD. $\endgroup$ Nov 15, 2016 at 11:17
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Let $A^TAv = \mu_k v$, $v\ne 0$. My idea was to use a vector of the form $\pmatrix{ v \\ Av}$, then $Bv$ is of the same shape.

With scalars $a,b$ $$ B \pmatrix{ av \\ bAv} = \pmatrix{ bA^TAv\\ aAv} = \pmatrix{ \mu_k b v\\ a Av}. $$ In order to obtain eigenvectors of $B$ to some eigenvalue $\lambda$, it must hold $\mu_k b=\lambda a$ and $a = \lambda b$.

The choice $b=0$ is impossible: this would imply $a\ne 0$ (to obtain non-zero eigenvector) and $\lambda=0$, a contradiction to $a=\lambda b$.

Thus, we can set $b=1$, $a=\lambda$, $\mu_k = \lambda^2$. This shows that the vector $$ \pmatrix{ \pm \sqrt \mu_k v\\Av} $$ is an eigenvector of $B$ to the eigenvalue $\pm \sqrt\mu_k$.

This construction yields all eigenvectors to non-zero eigenvalues of $B$. Since the ranks of $A$, $A^T$, $A^TA$ are the same, one can construct eigenvectors of $B$ to the eigenvalue zero by $$ \pmatrix{v\\0}, \pmatrix{0\\w} $$ with $Av=0$, $A^Tw=0$. Thus it is possible to construct a basis of eigenvectors of $B$ with the help of the eigenbasis of $A^TA$ plus a basis of $\ker A^T$.

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