The course has met invertibility again and again in different clothes: a non-zero determinant, a pivot in every column, independent columns, a map with nothing in its kernel. For a square matrix these are all the same condition, and one of them holding makes every one of them hold.

A key that turns a lock can be described by its shape, by the lock it opens or by the door that swings. Every description picks out the same key.

For an $n\times n$ matrix $A$ with map $T$ on $\mathbb{R}^n$, the statements below are equivalent: if one of them holds, all of them do, and if one fails, all fail. Most were proved one at a time in earlier chapters. The one step still owed, that a one-sided inverse of a square matrix is an inverse, is the theorem below, and rank–nullity is what proves it.

• $A$ is invertible; $\det A\neq0$; $A^{T}$ is invertible.
• has pivots; its reduced form is ; .

From $BA=I$, a vector with $A\mathbf{x}=\mathbf{0}$ satisfies $\mathbf{x}=BA\mathbf{x}=\mathbf{0}$, so the kernel of is zero. Rank–nullity gives , and a square matrix with pivots is invertible. Then , so too. With the roles swapped, makes a left inverse of , and the same argument applies.

Multiply by $B$ on the left and use $BA=I$: only $\mathbf{0}$ is sent to $\mathbf{0}$.

Rank–nullity: $n=0+\operatorname{rank}A$.

A square matrix with $n$ pivots reduces to $I$, so it is invertible: the chapter on elimination.

Regroup the product: $B$ is the inverse of $A$.

So $B$ undoes $A$ from both sides.

To decide whether a square matrix is invertible, check whichever condition is easiest: a determinant, a count of pivots, or a relation among the columns. The answer is the same whichever is used.

• For a square $A$: invertible, $\det A\neq0$, $n$ pivots, rank $n$, independent columns, spanning columns, one-to-one, onto, all say the same thing.
• A one-sided inverse of a square matrix is an inverse.

Great work! You now know how to:

• decide invertibility by whichever condition is easiest to check;
• prove that a one-sided inverse of a square matrix is an inverse;
• use $\det(AB)=\det A\det B$ and know where it comes from;
• see why none of this carries over to rectangular matrices.