Proposition: Let $A$ be a $(n \times n)$-matrix. If the columns of $A$ are linearly dependent, then $\det(A) = 0$.
Attempt at proof: Let $A = (A_1, A_2, \ldots, A_n)$, where each $A_i$ is a column vector. Since the columns are linearly dependent, we have \begin{align*} \lambda_1 A_1 + \ldots + \lambda_n A_n = 0 \end{align*} where not all $\lambda_i$ are zero. Suppose, without loss of generality, that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = - \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n. \end{align*} It follows that \begin{align*} \det(A_1, \ldots, A_n) &= \det (- \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n; A_2, A_3, \ldots, A_n) \\ &= \sum_{i=2}^n - \frac{\lambda_i}{\lambda_1} \det(A_2 + \ldots + A_k + \ldots + A_n; A_2, A_3, \ldots, A_n) \end{align*} This follows since the determinant is linear in each column. Now, I want to write \begin{align*} = \sum_{i=2}^n - \frac{\lambda_i}{\lambda_1} \det(A_k, \ldots, A_k, \ldots) = 0 \end{align*} since two columns are equal. But I'm not sure if my notation is correct in the last step? How do I break down the $A_2 + \ldots + A_k + \ldots$ etc.? Suggestions are appreciated!