If $Ax=B$ has two solution, then there must be a third one?

How do I prove this conjecture?

Let $A$ be a matrix, and $B$ be a column vectore. If $Ax=B$ has two solutions, then there must be a third one.

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One of the theorems you already know about the number of solutions to a system of equations can be applied.... –  Hurkyl Apr 13 '13 at 14:44
Given any two vectors $x,y$ in a vector space over an infinite field $F$, their span $\mbox{Span}\;\{x,y\}=Fx+Fy$ is infinite as soon as one of them is nonzero. –  1015 Apr 13 '13 at 15:04

You can do even better than that, assuming your field is infinite. Suppose that $Ax_1=b$ and $Ax_2=b$, where $x_1\ne x_2$. Now, for any $t$ in your scalar field, define $y_t:=tx_1+(1-t)x_2$. Now, $y_1=x_1$ and $y_0=x_2$, and generally, for $s\ne t$ we have $y_s\ne y_t$ (why?). What is $Ay_t$? In this way, we see that we have infinitely many solutions, so long as the scalar field is infinite.

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Sorry... I feel dumb on this one. How is, for instance, $Ay_3$ a solution? –  ohad Apr 13 '13 at 15:42
Don't forget that $A(x+y)=Ax+Ay$ for any (appropriately-sized) column vectors $x,y,$ and that $A(cx)=c(Ax)$ for any scalar $c$ and any (appropriately-sized) column vector $x$. –  Cameron Buie Apr 13 '13 at 16:09
In your particular example, $A(3x_1)=3b$ and $A(-2x_2)=-2b$, whence $Ay_3=3b+-2b=b.$ –  Cameron Buie Apr 13 '13 at 16:24
Thank you. I just fail to understand why this happens really. I also fail using the @ function. Thanks again!!! –  ohad Apr 13 '13 at 17:02
When you comment on a question or answer, the original poster is notified automatically, so there's no need to use the @ in that case. As for why matrix multiplication distributes over vector addition, and why we can pull scalars out front, that's just a nice consequence of the way matrix multiplication is defined. For example, suppose $A$ is an $m\times n$ matrix with $a_{ij}$ indicating the entry of $A$ in the $i$th row, $j$th column. Let $x$ and $y$ be $n$-dimensional column vectors, with respective entries $x_1,\dots,x_n$ and $y_1,\dots,y_n$. Let $c$ be a scalar. (cont'd). –  Cameron Buie Apr 13 '13 at 18:39
Suppose $x_1$ and $x_2$ are two (different) solutions, namely $Ax_1=B,Ax_2=B$. Then $A(x_2-x_1)=0$. Let $r$ be the nullity of $A$ and $y_1,\cdots,y_r$ be linearly independent solutions of $Ay=0$. Then $$x=x_1+c_1y_1+\cdots+c_ry_r$$ is the general solution of $Ax=B$. Thus there are infinite many solutions (of course, there is a third one).