# Prove that the Matrix $BC$ has rank of at most $1$

$B$ is a $3\times1$ matrix and $C$ is a $1 \times 3$ matrix. Prove that the $3\times 3$ matrix $BC$ has rank at most $1$.

I view this in the form of linear map. $B \equiv L_{B} : F^1 \rightarrow F^3$

Hence rank($BC$) $\leq\$rank($B$) = rank($L_{B}$) $\leq$ $3$.

How do I conclude the stronger inequality of $\leq 1$ here ?

Any help or insight is deeply appreciated

• $$rank(B) \leq 1$$ ;) – N. S. Apr 11 '16 at 18:13

Note that $\ker C \subset \ker BC$ and hence, by rank-nullity, $\operatorname{rank} BC \leq \operatorname{rank} C \leq 1$.
Define the transformation corresponding to the matrix $BC$, i.e. $T(x)=BCx$, for every $3\times 1$ vector $x$. Since $Cx$ is an scalar always all the time, the image of the defined transformation is always an scalar of $B$, which means that the rank of $T$ and consequently the rank of $BC$ is at most one.