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If A is singular then it isn't invertible, so by the invertible matrix theorem the columns of A are not linearly independent.How would I prove that if A is singular, then Av=0 has a non-zero solution?.

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How can I prove that the columns of an invertible matrix are linearly independent (from 'first principles')?If A is singular then it isn't invertible, so by the invertible matrix theorem the columns of A are not linearly independent.

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Many of these proofs also work by proving a couple of statements and then using that to imply the other statements. Any true statement of the IMT implies all of the others so there are lots of ways to go between these ideas.

Here is an example of an answer to your question:

"Assume that for the matrix A, Row i = Row j. By interchanging these two rows, the determinant changes sign (by Property 2). However, since these two rows are the same, interchanging them obviously leaves the matrix and, therefore, the determinant unchanged. Since 0 is the only number which equals its own opposite, det A = 0"

This uses the property that switching two rows of a matrix will reverse the sign of the determinant.

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Anyway, that's all I have to offer since I don't know the way you want to approach it but I know that a handful of members here are very knowledgable of linear algebra so hopefully one of them can comment further.