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Suppose $A=uv^T$ where $u$ and $v$ are non-zero column vectors in ${\mathbb R}^n$, $n\geq 3$. $\lambda=0$ is an eigenvalue of $A$ since $A$ is not of full rank. $\lambda=v^Tu$ is also an eigenvalue of $A$ since $$Au = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Here is my question:

Are there any other eigenvalues of $A$?


Thanks to Didier's comment and anon's answer, $A$ can not have other eigenvalues than $0$ and $v^Tu$. I would like to update the question:

Can $A$ be diagonalizable?

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$[0,1;0,0]$ is rank $1$ but, two eigenvalues are zero. You are actually saying, rank $r$ means $r$ singular values are nonzero... –  Sunni Aug 2 '11 at 17:45
As to the last question: anon's answer shows that if $v^Tu\neq 0$, then the algebraic and geometric multiplicities of $\lambda=0$ are equal (both are $n-1$); and therefore $A$ is diagonalizable (the other eigenvalue has algebraic multiplicity $1$, so it poses no obstacle to diagonalizability). If $v^Tu=0$, on the other hand, then $A$ is diagonalizable if and only if $A$ is the zero matrix. –  Arturo Magidin Aug 2 '11 at 18:00
@Arturo: +1. Thanks. Now things are clear. –  Jack Aug 2 '11 at 18:05
@Jack: Actually, $A$ can never be the zero matrix under the assumption that neither $u$ nor $v$ are the zero vector. I've expanded the comment to show that $A$ is diagonalizable if and only if $u$ is not orthogonal to $v$. –  Arturo Magidin Aug 2 '11 at 18:13
@Arturo: Thanks for the further explanation. :) –  Jack Aug 2 '11 at 18:27
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3 Answers

up vote 6 down vote accepted

We're assuming $v\ne 0$. The orthogonal complement of the linear subspace generated by $v$ (i.e. the set of all vectors orthogonal to $v$) is therefore $(n-1)$-dimensional. Let $\phi_1,\dots,\phi_{n-1}$ be a basis for this space. Then they are linearly independent and $uv^T \phi_i = (v\cdot\phi_i)u=0 $. Thus the the eigenvalue $0$ has multiplicity $n-1$, and there are no other eigenvalues besides it and $v\cdot u$.

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As to your last question, when is $A$ diagonalizable?

If $v^Tu\neq 0$, then from anon's answer you know the algebraic multiplicity of $\lambda$ is at least $n-1$, and from your previous work you know $\lambda=v^Tu\neq 0$ is an eigenvalue; together, that gives you at least $n$ eigenvalues (counting multiplicity); since the geometric and algebraic multiplicities of $\lambda=0$ are equal, and the other eigenvalue has algebraic multiplicity $1$, it follows that $A$ is diagonalizable in this case.

If $v^Tu=0$, on the other hand, then the above argument does not hold. But if $\mathbf{x}$ is nonzero, then you have $A\mathbf{x} = (uv^T)\mathbf{x} = u(v^T\mathbf{x}) = (v\cdot \mathbf{x})u$; if this is a multiple of $\mathbf{x}$, $(v\cdot\mathbf{x})u = \mu\mathbf{x}$, then either $\mu=0$, in which case $v\cdot\mathbf{x}=0$, so $\mathbf{x}$ is in the orthogonal complement of $v$; or else $\mu\neq 0$, in which case $v\cdot \mathbf{x} = v\cdot\left(\frac{v\cdot\mathbf{x}}{\mu}\right)u = \left(\frac{v\cdot\mathbf{x}}{\mu}\right)(v\cdot u) = 0$, and again $\mathbf{x}$ lies in the orthogonal complement of $v$; that is, the only eigenvectors lie in the orthogonal complement of $v$, and the only eigenvalue is $0$. This means the eigenspace is of dimension $n-1$, and therefore the geometric multiplicity of $0$ is strictly smaller than its algebraic multiplicity, so $A$ is not diagonalizable.

In summary, $A$ is diagonalizable if and only if $v^Tu\neq 0$, if and only if $u$ is not orthogonal to $v$.

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For the special case where $u$ and $v$ are orthogonal, it suffices to note that since $v^T u \neq 0$, it cannot be diagonalizable with all its eigenvalues equal to zero. –  Ryan Reich Dec 6 '13 at 20:49
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The matrix $A=uv^T$ has rank$~1$, unless either $u$ or $v$ is zero, in which case $A=0$; assume the latter is not the case. By rank-nullity, $\ker(A)$ (the eigenspace of$~A$ for the eigenvalue$~0$) has dimension $n-1$, so $0$ is a root of the characteristic polynomial $\chi_A$ with multiplicity at least$~n-1$. The sum of all such roots (counted with multiplicity) is $\def\tr{\operatorname{tr}}\tr A=v^T u$, so $\chi_A=x^{n-1}(x-\tr A)$. If $\tr A\neq0$ then the (direct) sum of the eigenspaces has dimension$~n$ and $A$ is diagonalisable; if however $\tr A=0$ then $0$ is the unique eigenvalue, but we assumed $A\neq0$, so $\dim\ker(A)<n$ and $A$ is not diagonalisable in this case.

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