Rank & Nullity (Matrices)

Think of an $m\times n$ matrix $A$ as a machine that takes an input vector $\mathbf{x}\in\mathbb{R}^{n}$ and produces an output $A\mathbf{x}\in\mathbb{R}^{m}$. Here $m$ is the number of rows and $n$ is the number of columns.

The output $A\mathbf{x}$ is always a combination of the columns of $A$: $A\mathbf{x}=x_1\,\mathbf{a}_1+x_2\,\mathbf{a}_2+\cdots+x_n\,\mathbf{a}_n,$ where $\mathbf{a}_j$ is the $j$-th column and $x_j$ is the $j$-th entry of $\mathbf{x}$.

• The set of all outputs $\{A\mathbf{x}\}$ is the column space $\operatorname{Col}(A)$.
• The rank of $A$ is the dimension of that column space: the number of genuinely independent directions the matrix can reach.

If some columns are redundant (a column is a combination of the others), they add no new direction, so the rank is smaller than $n$. Rank measures how much "spread" the matrix has in its output.

To measure rank concretely we use row reduction. Every matrix can be brought by elementary row operations to a unique reduced row echelon form (RREF), which looks like a staircase of leading $1$s.

• A pivot is a leading $1$ in a nonzero row of the RREF; the column it sits in is a pivot column.
• A free column is a column of the RREF with no pivot in it.

Two facts make this useful:

1. Row operations do not change the row space, so the number of nonzero rows in RREF equals $\dim(\text{row space})$.
2. Row operations preserve the dependence relations among columns, so a column of $A$ is independent of the earlier ones exactly when it becomes a pivot column.

Consequently the number of pivots is the fundamental count: it equals both the number of independent rows and the number of independent columns.

Let $A$ be an $m\times n$ matrix and let $R$ be its RREF.

Rank. $\operatorname{rank}(A)=\#\{\text{pivots in }R\}=\dim\operatorname{Col}(A)=\dim\operatorname{Row}(A).$

Null space. The null space is $\operatorname{Null}(A)=\{\mathbf{x}\in\mathbb{R}^{n}:A\mathbf{x}=\mathbf{0}\}$. When you solve $A\mathbf{x}=\mathbf{0}$, each free column corresponds to a free variable you can choose arbitrarily.

Nullity. $\operatorname{nullity}(A)=\dim\operatorname{Null}(A)=\#\{\text{free columns in }R\}=\#\{\text{free variables in }A\mathbf{x}=\mathbf{0}\}.$

So rank counts the pivot columns and nullity counts the free columns. Since every one of the $n$ columns is either a pivot column or a free column (never both, never neither), the two counts must add up to $n$.

Because each of the $n$ columns is either a pivot column or a free column, we get the central identity: $\boxed{\operatorname{rank}(A)+\operatorname{nullity}(A)=n}\qquad(n=\#\text{columns}).$

Reading it two ways.

• Know the matrix size and the rank? Then $\operatorname{nullity}(A)=n-\operatorname{rank}(A)$ with no extra work.
• Know the nullity instead? Then $\operatorname{rank}(A)=n-\operatorname{nullity}(A)$.

Useful bounds. Rank cannot exceed the number of rows or columns, so $0\le\operatorname{rank}(A)\le\min(m,n),\qquad\text{hence}\qquad \operatorname{nullity}(A)=n-\operatorname{rank}(A)\ge n-\min(m,n).$ A square $n\times n$ matrix is invertible exactly when $\operatorname{rank}(A)=n$, i.e. when $\operatorname{nullity}(A)=0$ (the only solution of $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$).

Warning. The $n$ on the right-hand side is the number of columns, not rows. A common mistake is to add rank and nullity to the number of rows.