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Suppose that $\tau \neq 0$，and vector $u\neq 0,v\neq 0\in R^n$, I want to find the eigenvalues of the matrix $E(u,v,\tau)=I-\tau uv^T$,here both $u$ and $v$ are column vectors,I have shown that $\lambda I-I-\tau uv^T=(\lambda -1)I-\tau uv^T$ and I need to show the principal minors of the matrix $\tau uv^T$ but I don't know how to calculate the higher order principal minors of thie matrix

YuerCauchy
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  • calculate the eigenvalues directly. Assuming $u,v$ are not dependent, given a vector $w$ that is orthogonal to $u,v,$ the result $Ew$ is especially simple. There are two more eigenvalues, one direction is $u,$ so what is $Eu?$ If you like, take unit $u/ |u| $ and then a perpendicular vector in the $u,v$ plane – Will Jagy Oct 10 '21 at 01:09
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    https://math.stackexchange.com/q/55165/321264 – StubbornAtom Oct 10 '21 at 08:09

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Concentrate on the matrix ${uv}^T$, if $w$ is any vector perpendicular to $v$ (these belong to an $(n-1)$ dimensional space), then $ {uv}^T w = 0$, and if $w$ is along $u$ then ${uv}^T w = (u^T v) u $, thus $u v^T$ has $(n-1)$ zero eigenvalues, and $1$ eigenvalue of $ u^T v $. Therefore, the eigenvalues of $(I - \tau u v^T)$ are $(n-1)$ eigenvalues each being equal to $1$, and one eigenvalue equal to $(1 - \tau (u^T v)) $

Hosam Hajeer
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