[proofplan] We prove the cycle (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (1). For (1) $\Rightarrow$ (2), [Determinant Multiplicativity](/theorems/395) gives $(\det A)(\det A^{-1}) = 1$, forcing $\det A \neq 0$. For (2) $\Rightarrow$ (3), we prove the contrapositive: rank less than $n$ gives linearly dependent columns, and the alternating property forces $\det A = 0$. For (3) $\Rightarrow$ (1), full rank means surjectivity, and [Injective iff Surjective in Finite Dimensions](/theorems/386) upgrades this to invertibility. [/proofplan] [step:Prove invertibility implies $\det A \neq 0$ via multiplicativity] Suppose $A$ is invertible with inverse $A^{-1}$. By [Determinant Multiplicativity](/theorems/395): \begin{align*} (\det A)(\det A^{-1}) = \det(AA^{-1}) = \det I_n = 1. \end{align*} Therefore $\det A$ has a multiplicative inverse in $\mathbb{F}$, so $\det A \neq 0$. [/step] [step:Prove $\det A \neq 0$ implies full rank via the contrapositive] We prove the contrapositive: if $\mathrm{rank}(A) < n$, then $\det A = 0$. If $\mathrm{rank}(A) < n$, the columns $A^{(1)}, \dots, A^{(n)}$ are linearly dependent: there exist $\lambda_1, \dots, \lambda_n \in \mathbb{F}$, not all zero, with $\sum_{j=1}^n \lambda_j A^{(j)} = \mathbf{0}$. Choose $k$ with $\lambda_k \neq 0$. Then \begin{align*} A^{(k)} = -\lambda_k^{-1} \sum_{j \neq k} \lambda_j A^{(j)}. \end{align*} By multilinearity of the determinant in the $k$th column: \begin{align*} \det A = -\lambda_k^{-1} \sum_{j \neq k} \lambda_j \det(A^{(1)}, \dots, A^{(j)}, \dots, A^{(n)}), \end{align*} where $A^{(j)}$ appears in both position $j$ and position $k$. Since the determinant is alternating, each term has two equal columns and hence equals zero. Therefore $\det A = 0$. [guided] We use the contrapositive: assume $\mathrm{rank}(A) < n$ and show $\det A = 0$. If $\mathrm{rank}(A) < n$, the $n$ columns of $A$ are linearly dependent in $\mathbb{F}^n$, meaning some column is a linear combination of the others. Pick $k$ such that $\lambda_k \neq 0$ in the dependence relation $\sum_{j=1}^n \lambda_j A^{(j)} = \mathbf{0}$. Solve for the $k$th column: $A^{(k)} = -\lambda_k^{-1} \sum_{j \neq k} \lambda_j A^{(j)}$. Now substitute this expression into the determinant, using multilinearity in the $k$th column: \begin{align*} \det A = \det(A^{(1)}, \dots, \underbrace{-\lambda_k^{-1} \sum_{j \neq k} \lambda_j A^{(j)}}_{k\text{th position}}, \dots, A^{(n)}) = -\lambda_k^{-1} \sum_{j \neq k} \lambda_j \det(A^{(1)}, \dots, \underbrace{A^{(j)}}_{k\text{th position}}, \dots, A^{(n)}). \end{align*} In each term of this sum, the column $A^{(j)}$ appears both in its original position $j$ and in position $k$. Since $\det$ is alternating, a matrix with two equal columns has determinant zero. Every term vanishes, giving $\det A = 0$. [/guided] [/step] [step:Prove full rank implies invertibility via the finite-dimensional equivalence] If $\mathrm{rank}(A) = n$, the [linear map](/page/Linear%20Map) $\alpha: \mathbb{F}^n \to \mathbb{F}^n$ defined by $\alpha(x) = Ax$ has $r(\alpha) = n = \dim\mathbb{F}^n$, so $\mathrm{im}\,\alpha = \mathbb{F}^n$ and $\alpha$ is surjective. By [Injective iff Surjective in Finite Dimensions](/theorems/386), $\alpha$ is an isomorphism, so $A$ is invertible. [/step]