Proof that $\det(A) = 0$ implies linear dependence of columns of a matrix $A$ Let $A$ be an $n \times n$ matrix. How would you rigorously prove that $\det(A) = 0$ if and only if the columns of $A$ are linearly dependent?
 A: Here are two possibilities. One is geometric.  If the columns of a square matrix are linearly dependent, the image of the unit cube under the transformation has measure zero, so $\det(A) = 0$.
Here is another.  If two matrices are row equivalent, they both have zero or non-zero determinant. IF the columns of a matrix are linearly dependent, it is of less than full rank (Remember: column rank = row rank). Hence is is row equivalent to a matrix with a zero row (or column if you wish).  In either event, its determinant is zero.
A: Using the fact that $\det(A)=\text{"product of the eigenvalues of $A$"}$:
$$
\begin{gather}
\text{$A$ has linearly dependent columns}\\
\Updownarrow\\
\text{$Ax=0$ for some $x\neq 0$}\\
\Updownarrow\\
\text{0 is an eigenvalue of $A$}\\
\Updownarrow\\
\det(A)=0
\end{gather}
$$
This is not really elementary way how to prove that (using the properties of the determinant with respect to the elementary row/column operations and the row echelon form), it is one of the simplest proofs and still rigorous enough.
