Eigenvalues of an Outer Product

Let $u,v\in\mathbb{R}^n$ and $A=uv^{T}$ (outer product). What are the eigenvalues of $$A$$ ?

Answer

Since $A=uv^{T}$ has rank $$1$$ , we have that $$n-1$$ linearly independent vectors will be sent to zero. All those eigenvectors therefore have eigenvalue $$0$$ . The only possible nonzero eigenvalue is found by the inner product (produces a scalar)

A u = (uv^{T})u = u\,(v^{T}u),

so $$u$$ is an eigenvector with eigenvalue $v^{T}u$ .

Hence the eigenvalues are $v^{T}u$ and $$0$$ (with multiplicity $$n-1$$ ).

Notes and comments

Comment 1: Note that in general, if a matrix $n \times n$ has rank $$1$$ , then it has $$n-1$$ eigenvalues equal to $$0$$ , and we are left with finding the remaining eigenvalue.

Comment 2: The inner product $u^{T}v$ is a scalar, while the outer product $uv^{T}$ is an $n\times n$ matrix.