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Let $V^T$ denote the transpose of a vector $V$. Consider two non-zero $p$-dimensional column vectors $a$ and $b$, for $p\ge 2$.

How many non-zero distinct eigen-values does the $p\times p$ matrix $ab^T + b a^T$ have?


marked as duplicate by Arnaud D., Arnaud Mortier, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Leucippus Feb 10 '18 at 0:18

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  • $\begingroup$ For the formatting of formulas, please have a look here $\endgroup$ – 57Jimmy Feb 9 '18 at 15:42

Rearranging $(ab^T+ba^T)v$, where $v$ is an eigenvector, we get:

$$\begin{align} ab^Tv+ba^Tv &= a(b^Tv)+b(a^Tv) \\ &= a\langle b,v\rangle + b\langle a,v\rangle \\ &= \lambda v \end{align}$$

Which means that $v$ is of the form $v = v_aa + v_bb$. Now substitute that back into the previous equation. To get the eigenvalues, we need to solve:

$$ \begin{bmatrix} \langle a,b\rangle && \|b\|^2 \\ \|a\|^2 && \langle a,b\rangle \end{bmatrix} \begin{bmatrix} v_a \\ v_b \end{bmatrix} = \lambda\begin{bmatrix} v_a \\ v_b \end{bmatrix} $$ $$\begin{align} &\Rightarrow \left(\langle a,b\rangle - \lambda\right)^2 = \|a\|^2\|b\|^2 \\ &\Rightarrow \lambda = \langle a,b\rangle \pm\|a\|\|b\| \end{align}$$

That gives $2$ eigenvalues, unless $a$ and $b$ are parallel, in which case one of the $\lambda$ above will be zero.


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