Suppose A=uvT where u and v are non-zero column vectors in Rn, n≥3. 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 v≠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 ϕ1,…,ϕn−1 be a basis for this space. Then they are linearly independent and uvTϕi=(v⋅ϕi)u=0. Thus the the eigenvalue 0 has multiplicity n−1, and there are no other eigenvalues besides it and v⋅u.

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