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The determinant of a square matrix is zero if and only if the column vectors are linearly dependent.

I see a lot of references to this all over the web, but I can't find an actual explanation for this anywhere.

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What definition of determinant are you using? – KReiser Sep 20 '13 at 5:38
Do you know it is zero if you can find linearly dependent "rows" inside the matrix? – Babak S. Sep 20 '13 at 6:13
Hint:… – dls Sep 20 '13 at 6:24
@dls: The answer will be covered by the points in the link but the OP asked about columns, so he/she does not know that the ranks of spaces which rows and columns create are the same. – Babak S. Sep 20 '13 at 6:30
Well, $\det$ is multilinear and anti-symmetric. Done. – Michael Hoppe Sep 20 '13 at 9:02

2 Answers 2

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The reason why is because if you have a matrix whose column vectors are linearly independent then when you look at reduced row echelon form there will end up being a zero row which means that parameter for the system can be any value you like. Meaning the system has infinitely many solutions. Also recall in reduced row echelon form the diagonal elements will be 1's excluding the row of zeros. Finally, the determinant of a upper triangular matrix is the product of the diagonal elements, therefore the determinant will be zero. It would look something like $$ A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c\\ 0& 0 & 0 \end{pmatrix}. $$ $Det(A) = 1\times1\times0 = 0$.

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Do you know that adding a multiple of one column to another column does not change the determinant? Do you see that if the columns are linearly dependent, then there is a way of adding multiples of columns to other columns so that one column becomes all zeros? Do you know what you can say about the determinant of a matrix that has an all-zero column?

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