# Eigenvalues of the rank one matrix $uv^T$

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?

• $[0,1;0,0]$ is rank $1$ but, two eigenvalues are zero. You are actually saying, rank $r$ means $r$ singular values are nonzero... Commented Aug 2, 2011 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. Commented Aug 2, 2011 at 18:00
• @Arturo: +1. Thanks. Now things are clear.
– user9464
Commented Aug 2, 2011 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 $v$. Commented Aug 2, 2011 at 18:13
• @Arturo: Thanks for the further explanation. :)
– user9464
Commented Aug 2, 2011 at 18:27

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 $v\cdot u$.

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 $v$.

• 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. Commented Dec 6, 2013 at 20:49
• In the case where $v^Tu=0$, and $\mu \neq 0$, isn't this a contradiction, because doesn't this imply that $\mu \neq 0$ is a non-zero eigenvalue, but with a corresponding eigenvector that is in the orthogonal complement of $v$? But all vectors in orth. comp. of $v$ correspond to eigenval $0$, no? Commented Jan 19, 2018 at 7:47
• The contradiction still results in the same conclusion, that when $v^Tu=0$, the only eigenvectors are in the orthogonal complement. But i just wanted to make sure I'm following all the logical steps. Commented Jan 19, 2018 at 7:48
• @mathiness: Yes, I think you could argue that way as well. Commented Jan 19, 2018 at 21:33

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 $$\lambda=0$$ 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 $$(n-1)+1=n$$, which implies that $$A$$ is diagonalisable. If however $$\tr A=0$$ then $$0$$ is the unique eigenvalue; but its eigenspace being of dimension only $$n-1$$, this means that $$A$$ is not diagonalisable in this case.

Another way to see that there are no eigenvalues other than $$0$$ and $$c=v^T u$$ is by doing the computation $$(A-cI)\circ A=uv^Tuv^T-(v^Tu)uv^T=0$$ since $$v^T u$$ is scalar. So eigenvalues of $$A$$ are roots of the polynomial $$(X-c)X$$; indeed the is the minimal polynomial of$$~A$$ (see this related answer).