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Find eigenvalues and rank of the matrix $A=I+xy^T$, where $x,y\in\mathbb R^n$ and $x^Ty=0$.

I have found a property that $\sum_{i=0}^n\lambda_i = \text{Tr}(A) = \text{Tr}(I+xy^T) = \text{Tr}(I) + \text{Tr}(xy^T) = n + \sum_{i=1}^nx_iy_i = n + x^Ty = n + 0 = n.$

But I am unsure of where to proceed from here.

Owen Murphy
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    If you know the eigenvalues of $xy^T$ then adding $I$ just shifts each one up by $1$. – podiki Nov 11 '21 at 23:23
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    $xy^T$ is rank 1, and let $\lambda$ be the possible "at most" one nonzero eigenvalue of $xy^T$, but $\lambda = Tr(A) = \sum_{i=1}^nx_iy_i = x^Ty = 0$ (using $x^Ty=0$ as in post). Then $\pi(\lambda) = \lambda ^n$ which tells us that 0 are the eigenvalues of $xy^T$. But how do I show what you have said @podiki? – Owen Murphy Nov 11 '21 at 23:26
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    It's true for any matrix $B$ that $A=B+I$ has eigenvalues of $B$ shifted up by $1$. Just use the same eigenvectors. $Bv=\lambda v$ means $Av = (\lambda+1)v$. – podiki Nov 11 '21 at 23:28
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    $xy^T$ being a rank 1 matrix has $0$ as a multiple value with multiplicity $(n-1)$; the remaining eigenvalue is therefore equal to the trace of $xy^T$... – Jean Marie Nov 11 '21 at 23:30
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    https://math.stackexchange.com/q/55165/321264, https://math.stackexchange.com/q/904926/321264 – StubbornAtom Nov 12 '21 at 10:34

1 Answers1

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First of all, if $a\neq 0$, then $(aI+xy^T)(\frac{1}{a}I-\frac{1}{a^2}xy^T)=I$ which implies $aI+xy^T$ is invertible making its determinant non zero. Taking $a=1$ reveals that $A$ has rank $n$.

The characteristic polynomial of $A$ is $\det\Big((1-\lambda)I+xy^T\Big)$ which doesn't stand a chance of equaling zero if $\lambda \neq 1$ from our aforementioned remark. On the other hand, if $\lambda=1$, the characteristic polynomial of $A$ evaluates to $\det(xy^T)$ which must equal zero as $xy^T$ has rank at most $1$.

In conclusion, the only eigenvalue of $A$ is $\lambda = 1$.

Matthew H.
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