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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2Note that $vv^{T}$ has rank one, as long as $v$ is not the zero vector. – Geoff Robinson Feb 04 '12 at 16:32
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@GeoffRobinson: Can you please provide more details? – littleEinstein Feb 04 '12 at 16:39
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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 $vv^T$ – Chris Janjigian Feb 04 '12 at 16:42
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2think 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 04 '12 at 16:44
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@littleEinstein: Others have by now provided the necessary details – Geoff Robinson Feb 04 '12 at 17:28
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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 $(vv^t)x = v(v^t x) = v(v \cdot x)=v0=\mathbf{0}=0x$.
Hans Lundmark
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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! – Ying Jan 04 '18 at 17:57
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1It'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. To find the eigenvectors, just solve the equation $\sum v_i x_i = 0$; it has a solution with $n-1$ parameters. – Hans Lundmark Jan 04 '18 at 20:14
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Sorry, why would this guarantee that the algebraic multiplicity and geometric multiplicity of eigenvalue 0 is the same? – user614287 May 03 '19 at 20:00
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1@mathpadawan: The geometric multiplicity of $\lambda=0$ is $n-1$ (the eigenspace has dimension $n-1$, as explained in my answer). And the algebraic multiplicity is always at least as large, so it must be $n-1$ too. (Neither multiplicity can be $n$, since there must also be room left for the other eigenvalue $\lambda=1$.) – Hans Lundmark May 03 '19 at 20:32
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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 $1$.
Davide Giraudo
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How to go from "rank 1" to eigenvalue multiplicity for eigenvalue 0? I see a gap here. – littleEinstein Feb 04 '12 at 17:49
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1It's rank nullity theorem: http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. – Davide Giraudo Feb 04 '12 at 17:54