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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.2012-02-04
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    @GeoffRobinson: Can you please provide more details?2012-02-04
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    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$2012-02-04
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    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 $v^t$.2012-02-04
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    @littleEinstein: Others have by now provided the necessary details2012-02-04

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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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    Why $v(v^t x)=v(v⋅x)$ is this true? Also, since the eigenvalue $0$ has multiplicity $n-1$, how to find $n-1$ linearly independent corresponding eigenvector? Thanks!2018-01-04
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    It's because $v^t x = v \cdot x = \sum v_i x_i$. Or, if you're pedantic enough to worry about the $1\times 1$ matrix $v^t x$ not being quite the same thing as the scalar $v \cdot x$, see [this answer](https://math.stackexchange.com/a/2422752/1242). To find the eigenvectors, just solve the equation $\sum v_i x_i = 0$; it has a solution with $n-1$ parameters.2018-01-04
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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.2012-02-04
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    It's rank nullity theorem: http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem.2012-02-04