Eigenvalues of the rank one matrix uvTuv^T

Suppose A=uvT where u and v are non-zero column vectors in Rn, n3. λ=0 is an eigenvalue of A since A is not of full rank. λ=vTu is also an eigenvalue of A since
Au=(uvT)u=u(vTu)=(vTu)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 vTu. I would like to update the question:

Can A be diagonalizable?

Answer

We’re assuming v0. The orthogonal complement of the linear subspace generated by v (i.e. the set of all vectors orthogonal to v) is therefore (n1)-dimensional. Let ϕ1,,ϕn1 be a basis for this space. Then they are linearly independent and uvTϕi=(vϕi)u=0. Thus the the eigenvalue 0 has multiplicity n1, and there are no other eigenvalues besides it and vu.

Attribution
Source : Link , Question Author : Community , Answer Author : anon

Leave a Comment