TITLE: Some basic questions regarding rank-1 matrices
QUESTION [6 upvotes]: If an $n\times n$ matrix $B$ has rank 1, and A is another $n\times n$ matrix, then why does $AB$ also have rank 1?  This showed up in a solution that I read through, but it doesn't seem like an obvious fact.
And one more thing that came up in this solution:  it says that since this matrix has rank 1, then it must have $(n-1)$ eigenvalues that are all zero, and only one non-zero eigenvalue.  I don't see how this has to be true either.
Any ideas are welcome.
Thanks,

REPLY [4 votes]: It comes from the associavity of matrix multiplication. If $B$ has rank-1 then it can be written in the form $B = u v^T$ for some vectors $u$ and $v$. So,
$$AB = A (uv^T) = (Au) v^T$$
but $Au$ is a vector itself, so now we have a rank-1 expression for $AB$. Maybe it will be helpful to see a picture of the shapes of the matrices during this process:
$$\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot\end{bmatrix}}_{A} \left(\underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{u}\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T}\right)= 
\left(\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot\end{bmatrix}}_{A}\underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{u}\right)\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T} =
\underbrace{\begin{bmatrix}\cdot \\ \cdot \\ \cdot\end{bmatrix}}_{Au}\underbrace{\begin{bmatrix}\cdot & \cdot & \cdot\end{bmatrix}}_{v^T} $$
(Of course, note that if $Au=0$, then the result is the zero matrix so the rank would be zero instead of 1)