Rank, Nullity and the Four Fundamental Spaces

Count pivots once, interpret them twice

For an $m\times n$ matrix $A$ , its rank $r$ is the number of pivots. It equals both the dimension of the column space and the dimension of the row space. The rank-nullity theorem states

\dim N(A)+\operatorname{rank}(A)=n.

There are $n-r$ free input directions that map to zero. Applying the same theorem to $A^T$ gives $\dim N(A^T)=m-r$ .

The four spaces are $C(A)\subseteq\mathbb R^m$ , $N(A)\subseteq\mathbb R^n$ , $C(A^T)\subseteq\mathbb R^n$ (row space), and $N(A^T)\subseteq\mathbb R^m$ (left null space). Their dimensions are $r,n-r,r,m-r$ respectively.

Visual guide

VISUAL GUIDEA rank-one map collapses one direction

Nullspace y = −x
Column space y = 0

The map A(x, y) = (x + y, 0) sends every output to the horizontal axis. Its nullspace is the line y = −x: all those inputs map to zero. Input dimension 2 splits into rank 1 plus nullity 1.

Worked example: count and construct

Let $A=\begin{pmatrix}1&2&3\\2&4&6\end{pmatrix}$ . Its rank is one. The column space is spanned by $(1,2)^T$ , and the row space by $(1,2,3)^T$ . The null space has basis $(-2,1,0)^T,(-3,0,1)^T$ . The left null space is spanned by $(-2,1)^T$ .

Check dimensions: the input has three coordinates, with one row-space direction and two null directions. The output has two coordinates, with one reachable direction and one left-null direction.

Worked example: a consistency certificate

A system $A\mathbf x=\mathbf b$ can be solvable only if every $\mathbf y\in N(A^T)$ satisfies $\mathbf y^T\mathbf b=0$ , because $\mathbf y^TA\mathbf x=0$ . This condition is also sufficient: the column space is the orthogonal complement of the left null space.

For the matrix above and $\mathbf b=(1,3)^T$ , the left-null vector $(-2,1)^T$ gives dot product $1\ne0$ . The system is inconsistent. For $\mathbf b=(1,2)^T$ , the test gives zero and the target is reachable.

Practice

A $4\times6$ matrix has rank three. Find both nullities.
Can a $3\times5$ matrix be one-to-one on $\mathbb R^5$ ?
When is a square $n\times n$ matrix invertible in terms of rank?

Show worked solutions

$\dim N(A)=3$ and $\dim N(A^T)=1$ .
No. Its rank is at most three, so its nullity is at least two.
Exactly when its rank is $n$ , equivalently its null space contains only zero.