[proofplan] We compare all five conditions through the linear endomorphism $T_A:k^n\to k^n$ represented by the matrix $A$. The [rank-nullity theorem](/theorems/916) shows that trivial kernel, full image, and full matrix rank are equivalent for an endomorphism of the $n$-dimensional [vector space](/page/Vector%20Space) $k^n$. The definition of the [general linear group](/page/General%20Linear%20Group) $GL_n(k)$ identifies membership in $GL_n(k)$ with invertibility of $T_A$, hence with bijectivity. Finally, the standard determinant criterion for square matrices over a field gives invertibility exactly when $\det A$ is nonzero. [/proofplan] [step:Relate trivial kernel, full image, and full rank by rank-nullity] Let $T_A:k^n \to k^n$ denote the $k$-[linear map](/page/Linear%20Map) $x\mapsto Ax$. Its nullity is $\dim_k\ker T_A$, and its matrix rank is \begin{align*} \operatorname{rank} A = \dim_k\operatorname{im} T_A \end{align*} By the rank-nullity theorem applied to the linear map $T_A:k^n\to k^n$, whose domain is finite-dimensional over $k$, \begin{align*} \dim_k\ker T_A + \operatorname{rank} A = n \end{align*} Therefore $\ker T_A=\{0\}$ if and only if $\dim_k\ker T_A=0$, if and only if $\operatorname{rank} A=n$. Since $\operatorname{im} T_A$ is a subspace of the $n$-dimensional vector space $k^n$, we have $\operatorname{im} T_A=k^n$ if and only if $\dim_k\operatorname{im} T_A=n$, that is, if and only if $\operatorname{rank} A=n$. Hence \begin{align*} \ker T_A=\{0\}\iff \operatorname{rank} A=n\iff \operatorname{im} T_A=k^n \end{align*} [guided] The point of this step is that the three linear-algebraic conditions are all measuring the same obstruction: whether $T_A$ loses dimension. We use the rank-nullity theorem for the map $T_A:k^n \to k^n$, whose domain has dimension $n$ over $k$. The theorem gives \begin{align*} \dim_k\ker T_A + \dim_k\operatorname{im} T_A = n \end{align*} By the definition of matrix rank for the linear map represented by $A$ in the standard ordered basis, \begin{align*} \operatorname{rank} A = \dim_k\operatorname{im} T_A \end{align*} Thus the rank-nullity formula becomes \begin{align*} \dim_k\ker T_A + \operatorname{rank} A = n \end{align*} Now $\ker T_A=\{0\}$ holds exactly when $\dim_k\ker T_A=0$. Substituting this into the displayed formula gives $\operatorname{rank} A=n$. Conversely, if $\operatorname{rank} A=n$, then the formula forces $\dim_k\ker T_A=0$, and the only zero-dimensional subspace is $\{0\}$. Hence \begin{align*} \ker T_A=\{0\}\iff \operatorname{rank} A=n \end{align*} It remains to connect rank with the image condition. Since $\operatorname{im} T_A$ is a subspace of $k^n$, and $k^n$ has dimension $n$, the equality $\operatorname{im} T_A=k^n$ holds exactly when $\dim_k\operatorname{im} T_A=n$. But $\dim_k\operatorname{im} T_A=\operatorname{rank} A$, so \begin{align*} \operatorname{im} T_A=k^n\iff \operatorname{rank} A=n \end{align*} Combining the two equivalences gives \begin{align*} \ker T_A=\{0\}\iff \operatorname{rank} A=n\iff \operatorname{im} T_A=k^n \end{align*} [/guided] [/step] [step:Identify membership in $GL_n(k)$ with bijectivity of the associated map] By the definition of the general linear group, $A\in GL_n(k)$ means that there exists a matrix $B\in M_n(k)$ such that \begin{align*} AB=I_n \end{align*} and \begin{align*} BA=I_n \end{align*} Here $I_n$ denotes the $n\times n$ identity matrix, and $\operatorname{id}_{k^n}$ denotes the identity map on $k^n$. Define the $k$-linear map \begin{align*} T_B:k^n\to k^n \end{align*} by $T_B(x)=Bx$. The identities $AB=I_n$ and $BA=I_n$ say that $T_A\circ T_B=\operatorname{id}_{k^n}$ and $T_B\circ T_A=\operatorname{id}_{k^n}$, because matrix multiplication represents composition of the associated linear maps in the standard ordered basis. Hence $T_A$ is bijective with inverse $T_B$. Conversely, if $T_A$ is bijective, then its inverse $T_A^{-1}:k^n\to k^n$ is a $k$-linear map. Let $B\in M_n(k)$ be the matrix of $T_A^{-1}$ with respect to the standard ordered basis of $k^n$, and define $T_B:k^n\to k^n$ by $T_B(x)=Bx$. Then $T_A\circ T_B=\operatorname{id}_{k^n}$ and $T_B\circ T_A=\operatorname{id}_{k^n}$, so $AB=I_n$ and $BA=I_n$. Therefore $A\in GL_n(k)$ if and only if $T_A$ is bijective. [/step] [step:Convert bijectivity into the kernel and image conditions] For any linear map, injectivity is equivalent to having zero kernel. Applied to $T_A$, this gives \begin{align*} T_A\text{ is injective}\iff \ker T_A=\{0\} \end{align*} For any map, surjectivity onto its codomain is equivalent to its image being the codomain. Applied to $T_A:k^n\to k^n$, this gives \begin{align*} T_A\text{ is surjective}\iff \operatorname{im} T_A=k^n \end{align*} Since $T_A$ is bijective if and only if it is both injective and surjective, and the previous step already showed that the kernel, image, and rank conditions are equivalent, we obtain \begin{align*} A\in GL_n(k)\iff \ker T_A=\{0\}\iff \operatorname{im} T_A=k^n\iff \operatorname{rank} A=n \end{align*} [/step] [step:Use the determinant criterion for invertibility] We use the standard determinant criterion for square matrices over a field: for $A\in M_n(k)$, \begin{align*} A\in GL_n(k)\iff \det A\ne 0 \end{align*} Its hypotheses apply because $A$ is an $n\times n$ matrix with entries in the field $k$. Combining this equivalence with the equivalences proved above gives \begin{align*} A\in GL_n(k)\iff \ker T_A=\{0\}\iff \operatorname{im} T_A=k^n\iff \operatorname{rank} A=n\iff \det A\ne 0 \end{align*} This is precisely the claimed equivalence of the five conditions. [/step]