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Given a column vector $v$ in $\mathbb{R}^n$, what are the eigenvalues of matrix $vv^T$ and associated eigenvectors?

PS: not homework even though it may look like so.

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Note that $vv^{T}$ has rank one, as long as $v$ is not the zero vector. – Geoff Robinson Feb 4 '12 at 16:32
@GeoffRobinson: Can you please provide more details? – littleEinstein Feb 4 '12 at 16:39
This follows from $rk(AB)\leq \min(rk(A),rk(B))$ (rk is rank) and the fact that if v is not the zero vector then something is in the image of $vv^T$ – Chris Janjigian Feb 4 '12 at 16:42
think about the rows of $vv^t$ where $v^t=(v_1,...,v_n)$. the first row is $v_1v^t$, the second row is $v_2v^t$, etc. so the rows are all multiples of $v^t$. – yoyo Feb 4 '12 at 16:44
@littleEinstein: Others have by now provided the necessary details – Geoff Robinson Feb 4 '12 at 17:28
up vote 4 down vote accepted

Assume $v \neq 0$. Then $v$ is an eigenvector with eigenvalue $|v|^2 >0$, since $(vv^t)v=v(v^t v)=v |v|^2 = |v|^2 v$, and any nonzero vector $x$ in the orthogonal complement of $v$ (which is of dimension $n-1$) is an eigenvector with eigenvalue zero, since $(vv^t)x = v(v^t x) = v(v \cdot x)=v0=\mathbf{0}=0x$.

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The columns of the matrix are $v_1v,\ldots,v_nv$ so if we take two column these one are linearly dependent, and so $vv^T$ has a rank of at most $1$. It's $0$ if $v=0$, and if $v\neq 0$, we have $\mathrm{Tr}A=|v|^2$ so the eignevalues are $0$ with multiplicity $n-1$ and $|v|^2$ with multiplicity $1$.

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How to go from "rank 1" to eigenvalue multiplicity for eigenvalue 0? I see a gap here. – littleEinstein Feb 4 '12 at 17:49
It's rank nullity theorem: – Davide Giraudo Feb 4 '12 at 17:54

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