How to show that each rank one decomposition of SVD is exactly rank one?

Question

by using the fact that rank(AB) $\leq$ min{rank(A), rank(B)}, I can only figure out that rank(u$\cdot$v$^{T}$) $\leq$ 1.

How to show that it is exactly rank one?

Appreciate any help!

Answer 1

You do not say so, but I assume that both $u$ and $v$ are column vectors.

The rank of a matrix is always a non-negative integer. From the fact that $\operatorname{rank}(uv^T) \leq 1$, you know that the rank is either $1$ or $0$. However, the only matrix with rank $0$ is the $0$-matrix. Show that if $uv^T = 0$, then either $u = 0$ or $v = 0$.