Proof of Rank of a Real Matrix: Rank–Nullity, Invertibility and the Size Bound

Routine facts about subspaces of $\mathbb{R}^{n}$ as vector spaces with the restricted operations are used without comment (Toolkit for AL3). The three clauses are proved in order; the second and third rest on the first.

By Properties of Dimension (vi), $\mathbb{R}^{n}$ is finite-dimensional with $\dim\mathbb{R}^{n}=n$. Since $L_{A}$ is linear (Real Matrices and Linear Maps between Euclidean Spaces (i)), The Rank–Nullity Theorem (i) applies to it: $\ker L_{A}$ and $\operatorname{im}L_{A}$ are finite-dimensional, and

$$n=\dim\ker L_{A}+\dim\operatorname{im}L_{A}.$$

As $\dim\operatorname{im}L_{A}=\operatorname{rank}A$ by The Rank of a Real Matrix (i), this is (i).

Now let $m=n$. By Real Matrices and Linear Maps between Euclidean Spaces (v), $A$ is invertible if and only if $\ker L_{A}=\{0\}$. The kernel is a vector space with the operations of $\mathbb{R}^{n}$ (Linear Subspaces Are Vector Spaces; Intersections (i)), finite-dimensional by the previous paragraph, so by Properties of Dimension (iv) it equals $\{0\}$ if and only if $\dim\ker L_{A}=0$. By (i), $\dim\ker L_{A}=0$ holds exactly when $\operatorname{rank}A=n$. This proves (ii).

Finally, $\operatorname{im}L_{A}$ is a subspace of $\mathbb{R}^{m}$ (Kernel and Image of a Linear Map Are Subspaces (ii)), so Properties of Dimension (v) and Properties of Dimension (vi) give $\operatorname{rank}A\le\dim\mathbb{R}^{m}=m$. On the other hand, (i) gives $\operatorname{rank}A=n-\dim\ker L_{A}\le n$. Hence

$$\operatorname{rank}A\le\min(m,n),$$

which is (iii).