## tacotime one year ago Linear Algebra: If A is a matrix and x and b are vectors, what is a necessary and sufficient constraint on A to ensure Ax=b has a non-zero solution for x?

WARNING: I AM REALLY TIRED RIGHT NOW. I AM NOT SURE WHETHER THE FOLLOWING IS CORRECT. It is clearly necessary for $$\underline{b}$$ to be non-zero in order for $$\underline{x}$$ to be non-zero. If $$\underline{b}=\underline{0}$$, then $$\underline{x}=\underline{0}$$ will be a solution to this equation. Let $$A$$ be a mxn matrix and suppose that $$b\in V$$, it seemed to me that $$\operatorname{rank}(A)=\dim(V)=m$$ is both necessary and sufficient for $$A\underline{x}=\underline{b}$$ to have a non-zero solution given $$\underline{b}\neq\underline{0}$$. If $$\operatorname{rank}(A)<\dim(V)$$, it will not span the entire $$V$$ and some $$\underline{v}\in V$$ will not have at least one $$\underline{x}$$ corresponding to it. $$\operatorname{rank}(A)>\dim(V)=m$$ seems contradictory to me. If $$n>m$$, then $$\operatorname{rank}(A)\leq m$$. If $$n<m$$, then $$\operatorname{rank}(A)\leq n<m$$. If $$n=m$$, then clearly $$\operatorname{rank}(A)\leq n=m$$. Neither way it will be greater than $$m$$.