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Linear algebra. Show that if the column of B are linearly dependent, then so are the columns of AB.
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Start with this...suppose \(A\) is \(m\times n\) and \(B\) is \(n\times k\) Let \(\mathbf{b}_1,\mathbf{b}_2,\ldots \mathbf{b}_k\) be the column vectors of \(B\) since they are L.D. then there is a nontrivial solution to \[c_1\mathbf{b}_1+c_2\mathbf{b}_2+\cdots+c_k\mathbf{b}_k=0\] write out what \(AB\) would be in terms of the column vectors of B...
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