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I want to show that the eigenvalues of $I - uv^T$, where $u,v \in \mathbb{R}^n$ are given by 1 with multiplicity $n-1$ and $1-v^Tu$ with multiplicity 1. I have tried setting up the eigenvalue equation $(I-uv^T)x = \lambda x$ which gives

$(1-\lambda)x - uv^Tx = 0$,

but I don't know how to proceed from here. Any help is appreciated!

Jim Totland
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  • Hint: $v^T x$ is a real number. Consider two cases: $x = \beta u$ and $x$ orthogonal to $v$ – Damien Nov 24 '20 at 16:05

2 Answers2

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First of all, the statement that you are trying to show is correct in the case that $v^Tu \neq 0$, but the case that $v^Tu = 0$ should be handled separately.

One approach to your problem in the case that $v^Tu \neq 0$ is to simply "guess" what the correct eigenvectors look like. In particular, I claim that if $x$ is orthogonal to $v$, then $x$ is an eigenvector. What is the associated eigenvalue? I also claim that if $x$ is parallel to $u$, then $x$ is an eigenvector. What is the associated eigenvalue?

Because $\operatorname{span}(u)$ and $\{v\}^\perp$ are complementary subspaces, we can then conclude that the operator $A = I - uv^T$ is diagonalizable and that all eigenvalues have been accounted for.

For the case that $v^Tu = 0$, it suffices to note that $[A - I]^2 = 0$. It follows that $A$ has $1$ as its only eigenvalue.

Ben Grossmann
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  • Thank you, I now understand how to find the eigenvalues, but I still struggle with how to determine their multiplicity. Any tips related to this? – Jim Totland Nov 24 '20 at 22:41
  • @JimTotland The geometric multiplicity of an eigenvalue is simply the dimension of its eigenspace. What's the dimension of $\operatorname{span}(u)$? What's the dimension of ${v}^\perp$? – Ben Grossmann Nov 24 '20 at 22:44
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    @Morad That was a typo. Hopefully it’s clear now – Ben Grossmann May 18 '21 at 04:48
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Hint: Start by finding the eigenvalues of $uv^T$, a matrix of rank $1$.

lhf
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