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

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?

• Multiply $Ae = \lambda e$ on the left by $y^T$ and consider associativity of the products involved.
– Paul
Commented Apr 25, 2019 at 8:10
• Every column of $A$ is a multiple of $x$, hence it has rank 1.
– amd
Commented Apr 25, 2019 at 8:19
• @amd Thanks. That is the answer I was looking for. I can't see why you did not write it as an answer. Commented Apr 26, 2019 at 5:54
• Seemed too short to me to be an answer (and will likely get flagged as too short by overzealous reviewers), but I’m happy to make it so.
– amd
Commented Apr 26, 2019 at 5:56
• @amd Simple answer is the best answer, right? Commented Apr 26, 2019 at 5:59

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 $$1$$.
• $xy^Tcx=(y^Tx)cx$, how ? Commented Apr 26, 2019 at 6:41
• Because $c$ and $y^Tx$ are scalars, so we can move them around however we like. Write $d = y^Tx$, then $xy^Tcx = c(xy^Tx) = c x d = dcx$. Commented Apr 26, 2019 at 6:42
• In case this seems strange, understand that the expression $y^Tx$ is playing two very distinct roles: First, a $1 \times 1$ matrix, and second, a scalar. It wouldn't make sense to write $(y^Tx)x$ if $y^T x$ were a $1 \times 1$ matrix, because the dimensions are incompatible with the column matrix $x$. Unfortunately, this cavalier notation is standard. Commented Apr 26, 2019 at 6:59
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, $$k=y^Tx$$.