condition to be full-rank with two matrices

Question

Let $A$ be $n \times m$ and $B$ be $n \times r$ with $m>r$. Consider a linear combination $Ax$ and $By$ where $x$ is $m \times 1$ and $y$ is $r \times 1$. Suppose that $Ax=By$ holds. In this case, if $B$ is of full rank, can we say that $A$ is also of full rank? If so, how can I prove it?

Answer 1

$\newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}}$

I understand your question in this way: Your $x$ and $y$ are fixed and you know that $Ax = By$ and $B$ has full rank. Does it follow that $A$ has full rank as well?

No. Denote by $\mathbb{F}$ the field over which your matrices are defined (if you work with integral domains instead the argument still holds). Let $r = n$. Choose $B = E_n\in\mathbb{F}^{n\times n}$ the identity matrix. This has full rank. Now choose some $A\in \mathbb{K}^{n\times m}$ which has not full rank and some $x\in \mathbb{F}^{m}$. Define $y := Ax \in \mathbb{F}^n$. Then

$$Ax = y = E_n y = By$$

so the condition is satisfied for those elements but $A$ does not have full rank.

Let's get more abstract. We can think of these matrices as linear maps $$A: \mathbb{F}^m \to \mathbb{F}^n$$

and

$$B: \mathbb{F}^r \to \mathbb{F}^n\,.$$ The condition $Ax = By$ (for $x, y \neq 0$) means that these linear maps have a non-trivial intersection of their images in $\mathbb{F}^n$. But well, the fact that $A$ and $B$ share some elements in the image will not transport some properties of $B$ to $A$. Full rank of $B$ means that $B$ is injective or surjective. And of course $B$ can be injective or surjective while $A$ isn't one of these things and they still share some non-trivial elements in the image. For example $$A\left(\mycolv{a\\b\\c\\d}\right) := \mycolv{a\\ 0}$$ is surly not injective and not surjective. But $$B\left(\mycolv{x\\y\\z}\right) := \mycolv{x\\y}$$ is surjective. Thus $B$ has full rank while $A$ has not. And we have $$A\left(\mycolv{a\\b\\c\\d}\right) = \mycolv{a\\ 0} = B\left(\mycolv{a\\0\\0}\right)\,.$$ The associated matrices would be

$$A = \begin{bmatrix}1 & 0 & 0 & 0\\0 & 0 & 0 &0\end{bmatrix}$$

and

$$B = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\end{bmatrix}$$ if you want to check this yourself with matrices.