If an $n \times n$ matrix $B$ has rank $1$, and $A$ is another $n \times n$ matrix, then why does $A B$ 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.
On
The rank of a matrix is the dimension of its image ${\rm rank} \ B = \dim \{Bx: x \in \mathbb R^n\},$ or equivalently the dimension of the space spanned by its columns.
Since applying $AB$ on $\mathbb R^n$ is the same as applying $A$ on the one-dimensional space $B\mathbb R^n$, we see that $AB$ can only have at most rank 1. Note that rank 0 is possible when the columns of $B$ are orthogonal to the rows of $A$.
We have the rank-nullity theorem $n = {\rm rank} \ B + \dim \ker B$ (i.e. what doesn't contribute to the dimensions of the image must land in the kernel.
If ${\rm rank} \ B = 1$, then the dimension of the kernel must be $n-1$, i.e. the eigenspace for the eigenvalue is $n-1$. We can deduce that there is also another nonzero eigenvalue since otherwise we'd have ${\rm rank} \ AB = 0$.
On
Concerning the first point, remember $A$ and $B$ represent endomorphisms $f$ and $g$ in some basis, and $AB$ represent $f\circ g$.
Now $\;\DeclareMathOperator\rk{rank}\DeclareMathOperator\img{Im}\rk B=1$ means $\;\dim\img g=1$. Then we know $\;\dim\img(f\circ g)\le\dim\img g$, which means $\;\rk AB\le\rk B$. So $\rk AB=0$ or $1$.
For the second point, if $AB$ has indeed rank $1$, its kernel has dimension $n-1$. As the kernel is the eigenspace for the eigenvalue $0$, and has dimension at most equal to the multiplicity of this eigenvalues, it implies the eigenvalue $0$ has multiplicity at least $n-1$, and it can't have multiplicity $n$ as it would imply $\rk AB=0$.
On
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)
Okay, we can show this in different ways, depending on how we define rank. I'll try to show these using both these definitions:
These two definitions are equivalent. The proof of this is left as an exercise to the reader.
If $B$ has rank 1, then $AB$ has at most rank one
Fixed the formulation for you on this one. $AB$ can have rank zero.
Definition 1
$B$ has rank one, so its range is one dimensional. It follows that its nullspace is $n-1$-dimensional. Consider the range of $AB$. If $x$ is in the nullspace of $B$, then $ABx = A0 = 0$, so $AB$'s nullspace is also at least $n-1$-dimensional, so its range is at most 1-dimensional.
Note that for $AB$ to have rank 1 you must have that $Ax \neq 0$ for some $x$ in the range of $B$. A sufficient, but not necessary, condition for this is that $A$ is invertible.
Definition 2
$B$ has rank one, so it can be written $B = xy^T$. Then $AB = Axy^T = (Ax)y^T$, so $AB$ can be written as one outer product (but we don't know if it can be written using zero outer products, which would be the case if $Ax = 0$).
A rank one matrix has $n-1$ zero eigenvalues
Definition 1
As said before, if $B$ has rank 1, then its nullspace is $n-1$ dimensional. Pick a basis $v_1, \dots, v_{n-1}$ for the nullspace of $B$. These are eigenvectors of $B$ with eigenvalue zero.
Definition 2
Say $B = xy^T$. Then if you multiply a vector $v$ with $B$ you get $xy^Tv$. Note that $y^Tv$ is the inner product of $y$ and $v$. The orthogonal complement $\mathcal U$ to the subspace spanned by $y$ has dimension $n-1$. If $u \in \mathcal U$ then $y^Tu = 0$ and hence $$Bu = xy^Tu = x \cdot 0 = 0$$ so if you pick a basis for $\mathcal U$, this basis will be eigenvectors to $B$ with eigenvalue 0.
Is there a non-zero eigenvalue?
In short, it is not guaranteed that there will be a non-zero eigenvalue.
This rank one nilpotent matrix: $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ has zero as eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. It therefore has $n-1$ zero eigenvalues, but it does not have 1 non-zero eigenvalue. In other words, it is not guaranteed that you will have a non-zero eigenvalue for rank one matrices.