# Rank of product of SVD vectors

Given a matrix $M$, we can compute its singular value decomposition $M=U\Sigma V^*$ where $^*$ is the complex conjugate transpose. $U$ and $V$ are unitary, so $UU^*=I$, $VV^*=I$. Let's take the $i$-th column of $U$, $u_i$. I've experimented with a few examples and I found that the matrix $u_iu_i^*$ always has rank 1. What theorem/property is behind this? Or even, is there a short and sweet proof of it?

## 1 Answer

Note that $u_i$ is $m\times1$ and $u_i^\ast$ is $1\times m$, so $u_iu_i^\ast$ is $m\times m$. Moreover, observe that the $k$th row of $u_iu_i^\ast$ is $u_i^\ast$ multiples the $k$th entry of $u_i$, for $1\le k\le m$. That is, each row of $u_iu_i^\ast$ is a scalar multiple of $u_i^\ast$, so we have ${\rm rank}(u_iu_i^\ast)\le1.$ On the other hand, since $u_i^\ast$ must be a non-zero vector, and at least one entry of $u_i$ is a non-zero scalar. So at least one row of $u_iu_i^\ast$ is non-zero, and then we have ${\rm rank}(u_iu_i^\ast)\ge1$. Hence ${\rm rank}(u_iu_i^\ast)=1$.

• Ah, so simple! Thanks a lot. – space_voyager May 8 '16 at 13:49