Eigenvalues of the rank one matrix uvTuv^T

Question 1

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 $A$ ?

Added:

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?

Question 2

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 $v\cdot u$ .

Eigenvalues of the rank one matrix uvTuv^T

Answer

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