If $AD+BD=CD$, is $A+B=C$?

Question

Assume non-square matrices $A, B, C \in \mathbb{R}^{n \times m}$ and $D \in \mathbb{R}^{m \times k}$. If we are given $$AD + BD = CD\,,$$ can we say that $$A+B=C \,?$$

I am thinking that we cannot just cross matrix $D$ out like we would do with a scalar, nor multiply with $D^{-1}$ since it's non-square.

What about the other way around? My guess is yes because we just multiply both sides with $D$. Any help is appreciated.

Answer 1

Your question is the same as asking if $$XA = YA \implies X=Y$$ is true and the answer is no. That would mean that map $X\mapsto XA$ which is linear, is injective i.e. it kernel is trivial which is not necessary.
Say $A= \pmatrix{0\;0\\0\;1}$ then $X=\pmatrix{1\;0\\0\;0\\0\;0} $ in it's kernel so it is not injective.

Or more direct example:

Say $$X=\pmatrix{1\;0\\0\;0\\0\;0}, Y=\pmatrix{0\;1\\0\;0\\0\;0},A= \pmatrix{0\;1\\0\;1}$$ then $$XA= \pmatrix{0\;1\\0\;0\\0\;0}=YA$$

Answer 2

Clearly $AD+BD=(A+B)D$ so we can waive about the sum and consider $$ AC =BC $$

As an example take $A,B$ to be row vectors ($1 \times m$) and C a column vector ($m \times 1$) .
Then $AC, BC$ represent the dot products of A with C and B with C.
From their equality, what you can tell is that they have the same projection on C, which means that they differ by a vector normal to $C$.

For The general case you can tell that the matrices differ by a matrix which belong to the null-space of C.