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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 nonzero solution for x?
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 nonzero solution for x?

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1WARNING: I AM REALLY TIRED RIGHT NOW. I AM NOT SURE WHETHER THE FOLLOWING IS CORRECT. It is clearly necessary for \(\underline{b}\) to be nonzero in order for \(\underline{x}\) to be nonzero. 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 nonzero 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 \(m\).
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