How to prove whether a matrix has rank $1$

Question

If $u$ $∈ \mathbb R^{m \times 1}$ and $v ∈ \mathbb R^{n \times 1}$ how do you show that the $(m \times n)$ matrix $uv^T$ has rank $1$? Would providing an example be sufficient to prove it?

Answer 1

Here is an intuitive explanation:

$U V^T$ will have a dimension of $m$ x $n$.

All the rows will be a linear combination(or multiples) of V i.e. Row Rank of the matrix is 1.

All columns will be a linear combination (or multiples) of U i.e. Column Rank of the matrix is 1.

Geometrically speaking, all the columns vectors of matrix will point in the same direction, along a line (1 -dimensional subspace, it is Subspace since the line passes through origin). That makes it Rank 1 matrix.

Answer 2

Just to add to the intuition : the outer product of $u$ and $v$ is given by $O=uv^T$. So, if $x$ is a vector and the $O$ acts on it, it will be given by $Ox=uv^Tx=u(v^Tx)$. As you can see $v^Tx$ is a dot product of $v$ and $x$ and is therefore a number. This number is the multiplied with $u$ which is a vector. So, whatever be the $x$ the output is a vector along $u$ which implies that the rank of the matrix (i.e., the dimension of the image space) must be $1$.

Answer 3

For general matrices $A$ and $B$ such that $AB$ is defined, it holds that $$ \operatorname{rank}AB\le\min\{\operatorname{rank}A,\operatorname{rank}B\} $$

The inequality $\operatorname{rank}AB\le\operatorname{rank}A$ is clear because the column space of $AB$ is contained in the column space of $A$.

The inequality $\operatorname{rank}AB\le\operatorname{rank}B$ follows from an application of the rank-nullity theorem. Suppose $A$ is $m\times n$ and $B$ is $n\times p$. Then clearly the null space of $B$ is contained in the null space of $AB$. Thus $\dim N(B)\le\dim N(AB)$. The rank-nullity theorem says that \begin{align} p&=\operatorname{rank}AB+\dim N(AB) \\ &=\operatorname{rank}B+\dim N(B) \end{align} so we get the desired inequality.

Also, the rank of an $m\times n$ matrix $A$ satisfies $$ \operatorname{rank}A\le\min\{m,n\} $$

In the particular case of $A=u$ and $B=v^T$, we have $$ \operatorname{rank}uv^T\le \min\{\operatorname{rank}u,\operatorname{rank}v^T\}\le 1 $$

If neither $u$ nor $v$ is zero, then $uv^T$ is nonzero, so its rank is at least $1$. If either $u$ or $v$ is zero, then the product $uv^T$ is the zero matrix, which has rank zero.

Note that also the converse holds: if an $m\times n$ matrix $A$ has rank $1$, then $A=uv^T$ for suitable column vectors $u$ and $v$.