Finding eigenvectors of $xy^T$

Question

Given vectors $x,y\in\mathbb{C}^n$, find all the eigenvectors and eigenvalues of the matrix $xy^\intercal$

I've already know that the eigenvalues are $0$ and $\mbox{trace}(xy^\intercal) = y^\intercal x$, but i don't know how to proceed for the eigenvectors asociated to $y^\intercal x$. For now i have

\begin{align} xy^\intercal v = (y^\intercal x)v \iff \begin{pmatrix}x_1y_1 -y^\intercal x &x_1y_2&\cdots&x_1y_n\\x_2y_1 & x_2y_2-y^\intercal x &\cdots&x_2y_n\\\vdots&&\ddots&\vdots\\x_ny_1&\cdots& & x_ny_n-y^\intercal x\end{pmatrix}v = \mathbf{0} \end{align}

Thanks in advance.

Because the trace of $xy^\intercal$ is an eigenvalue, so $v$ is an eigenvector iff $xy^\intercal v = y^\intercal x v \iff (xy^\intercal - y^{\intercal}xI_n)v = 0 $, that eliminates the diagonal — Raúl Astete, Apr 30 '20 at 22:45
@RaúlAstete It is not true that "that eliminates the diagonal" — Ben Grossmann, May 01 '20 at 00:27

Stefan Lafon · Accepted Answer · 2020-05-01T08:59:11.733

Suppose $u\neq 0$ is an eigenvector. Then by definition, there exists $\lambda\in\mathbb C$ such that $$(xy^T)u=x(y^T u)=\lambda u$$

Notice that $y^T u$ is a scalar (number). So that means that either $x$ and $u$ are co-linear, or $\lambda=0=y^T u$.

Conversely, it's easy to verify that $x$ itself is an eigenvector with eigenvalue $y^Tx$ (again, a number). And any vector $u$ that is orthogonal to $y$ is an eigenvector with eigenvalue $0$.

Finding eigenvectors of $xy^T$

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