If homogeneous system has infinitely many solutions, does every system with that left hand side?

Question

Hi I am trying to work on a proof question that I took note of because the answer did not come to mind immediately. I also found it was a good question to illustrate something I do not understand very well.

The question asks to prove, or disprove that if the system $AX=0$ has infinite solutions, then so to does the system $AX=B$ for any choice of $B$.

First off, I get confused by "For any choice of $B$". I thought $B$ represented the column of constants? How could you choose any $B$? Is that not implying that $B$ can be $(a,b,c,,,n)$ for any $a,b,c,,n$, in the real numbers?

Anyways, in regards to the proof. I know that if $AX=0$ has infinite solutions, then there must be atleast one parameter, thus $(n-r)\geq 1$ where $n$ is the number of variables and $r$ the rank of the matrix. I am not sure where to go from here.

Hopefully you guys can help me with this and get a better understanding!

Thank you,

Answer 1

Hint. Consider the system $$\begin{align}1x + 0y &= 1,\\1x + 0y &= 2.\end{align}$$

Answer 2

"For any choice of $B$" means that you need to prove the statement given an arbitrary vector $B$, meaning essentially what you said, $B=(a,b,c,\dotsc)$ for any choice of real numbers $a,b,c,\dotsc$.

My hint here is to use the distributive property of vector spaces, $A(X+Y)=AX+AY$. Take $X$ to be the solution of $AX=0$ and $Y$ to be the solution of $AY=B$. Now, you do the rest, and show that the converse holds, too.

Answer 3

What you are asked to prove or disprove is false in the following obvious way: there is no guarantee that for every choice of $B$ there are any solutions at all, let alone infinitely many of them. The most basic example is the $1\times1$ system with $A=0$: the "system" $0x=0$ has infinitely many solutions (all $x\in\Bbb R$) but for any other choice of $B$, for instance $B=1$ giving the equation $0x=1$ there are no solutions at all.

If you like you can blow this idea up a bit so that $A$ looks less trivial. Take for instance $A=(\begin{smallmatrix}1&-1\\2&-2\end{smallmatrix})$, then the system $$ \begin{align} x-\phantom2y&=0\\ 2x-2y&=0 \end{align} $$ has infinitely many solutions (namely any $x=y\in\Bbb R$), but for instance $$ \begin{align} x-\phantom2y&=1\\ 2x-2y&=5 \end{align} $$ has none, as you can easily see.