Invertible matrix theorem: invertibility, full pivot rank, RREF $I$, trivial nullspace and unique solvability are equivalent

By [L6], matrix invertibility is equivalent to the multiplication map being an isomorphism. An isomorphism has a two-sided inverse by [L8], hence is injective, and [L7] then gives $N(A)=\{0\}$.

By [L2], trivial nullspace is equivalent to rank $n$, and [L4] makes this equivalent to having $n$ pivots. A square matrix has $n$ pivots exactly when every row and column contains one.

An $n\times n$ RREF with a pivot in every row and column is $I_n$. By [L1], RREF $I_n$ is equivalent to the assertion that every right-hand side produces a consistent system with no free variable, hence exactly one solution.

Under condition 7, define $S(b)$ to be the unique solution of $Ax=b$. Then $L_A(S(b))=b$, and uniqueness for the right-hand side $Ax$ gives $S(L_A(x))=x$. Moreover, linearity of $L_A$ from [L3] gives $L_A(S(\lambda y+z))=\lambda y+z=L_A(\lambda S(y)+S(z))$; applying the inverse function $S$ proves that $S$ is linear. Thus $L_A$ has a two-sided linear inverse and is a linear isomorphism by [L8], so [L6] returns condition 1. Together with steps 1.1–3.1, this closes the implication cycle and proves all seven conditions equivalent. For $n=0$, the unique empty matrix is $I_0$, its map on the zero space is bijective, and every condition holds.