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Let $x$ and $y$ be two non-zero $n × 1$ vectors. If $y^T$ denotes the transpose of $y$, what are the eigenvalues of the $n × n$ matrix $xy^T$

I have only idea that given matrix has rank $1$.

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In your case, if $y \in \mathbb{R}^{n \times 1}$, there are $n-1$ mutually vectors perpendicular to $y$. These $n-1$ vectors are eigenvectors of the matrix $xy^T$ corresponding to the eigenvalue $0$. (In general, a $n \times n$ matrix of rank $r$ has $n-r$ eigenvalues as $0$.) The other eigenvalue is nothing but the trace of $xy^T$, which is nothing but $x^Ty$.