Singular Matrix and Linear Dependency

Feb 2015
I am asked to prove this: Matrix is singular<=>columns are linearly dependent<=>determinant is zero<=>matrix is singular
I have proved everything except for the first implication, which I can't find anywhere. Can someone prove that if a matrix is singular, its columns are linearly dependent?
Thanks in advance.

Country Boy

Math Team
Jan 2015
Multiply the given matrix by \(\displaystyle \begin{bmatrix}1 \\ 0 \\ \cdot\cdot\cdot \\ 0 \end{bmatrix}\), \(\displaystyle \begin{bmatrix} 0 \\ 1 \\ \cdot\cdot\cdot \\ 0\end{bmatrix}\), down to \(\displaystyle \begin{bmatrix} 0 \\ 0 \\ \cdot\cdot\cdot \\ 1\end{bmatrix}\). You should see that the resulting vector (column matrices) are the columns of the matrix. You should also be able to show that those vectors span the column space. Since the matrix is singular, those n vectors cannot be a basis for the column space and so must be dependent.