Properties of outer product of two unit vectors? Why is there only one non-zero eigenvalue for such a matrix?

Question

Let $x,y$ be two unit vectors.

$A=xy^T$ be the outer product.

The eigenvalues of A are seen to be $[0, 0, 0,...0, k]$. Why is that?

What are the properties of the outer product of two unit vectors?

Why is there only one non-zero eigenvalue for such a matrix?

Answer 1

Every column of $A$ is a multiple of $x$ and every row is a multiple of $y$. $x$ and $y$ are both nonzero, hence $A$ has rank 1. Moreover, $k=y^Tx$.

Answer 2

Here's an elementary proof. If $A = xy^T$ and $v$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda$, then $\lambda v = Av = (xy^T)v = x(y^T v)$. Note that $y^T v$ is a scalar. Assume $\lambda \ne 0$, and divide through by $\lambda$ to get $v = x(y^T v / \lambda)$. Put $c = y^Tv/\lambda$. We have $Av = A(cx) = xy^T cx = (y^Tx)cx = (y^T x)v$, so $\lambda = y^Tx$.

One way to show that the algebraic multiplicity of the eigenvalue $y^Tx$ is $1$ is to take the trace: $tr(A) = tr(xy^T) = tr(y^T x) = y^Tx$. The trace is the sum of eigenvalues, so if $y^Tx \ne 0$ then the multiplicity of $y^Tx$ must be $1$.