## The columns of a square matrix over a field are linearly independent if and only if the determiniant of the matrix is nonzero

Let $F$ be a field and let $A = [A_1 | \cdots | A_n]$ be a square matrix of dimension $n \times n$ over $F$. Prove that the set $\{A_i\}_{i=1}^n$ is linearly independent if and only if $\mathsf{det}\ A \neq 0$.

Let $B$ be the reduced row echelon form of $A$, and let $P$ be invertible such that $PA = B$.

Suppose the columns of $A$ are linearly independent. Now $B$ has column rank $n$. In particular, $B = I$. Now $1 = \mathsf{det}(PA) = \mathsf{det}(P) \mathsf{det}(A)$; so $\mathsf{det}(A) \neq 0$.

We prove the converse contrapositively. Suppose the columns of $A$ are linearly dependent; then the column rank of $B$ is strictly less than $n$, so that $B$ has a row of all zeros. Using the cofactor expansion formula, $\mathsf{det}(B) = 0$. Since $P$ is invertible, its determinant is nonzero; thus $\mathsf{det}(A) = 0$. Thus if $\mathsf{det}(A) \neq 0$, then the columns of $A$ are linearly independent.