Rank-Nullity Theorem

rank(A) + nullity(A) = n — every column counts exactly once.

The idea

Theorem (Rank–nullity theorem).

Let $A$ be an $m \times n$ matrix, and write $\text{rank}(A)$ for the dimension of its column space and $\text{nullity}(A)$ for the dimension of its null space. Then $\text{rank}(A) + \text{nullity}(A) = n,$ the number of columns.

The two numbers count opposite outcomes. The rank counts the independent directions that survive multiplication by $A$ and appear in the output — a direction $u$ of this kind keeps a nonzero image $Au$ in the column space; the nullity counts the independent input directions that $A$ sends to zero. The theorem says these counts exhaust the $n$ dimensions of the input space exactly: every dimension is either kept or destroyed, none is counted twice, and none goes missing.

Two consequences follow at once. Knowing either number gives the other. And although the theorem never mentions $m$, the rank of any matrix is at most $m$, so a wide matrix — more columns than rows — has nullity at least $n - m$: it must send some nonzero vector to zero.

Ways to work on it

• Walkthrough. Verify the rank–nullity theorem on simple matrices.
• Proof. Prove the rank–nullity theorem by extending a basis of the null space.
• Practice. Given the shape and rank, find the nullity.
• Hardest. Find the largest possible rank of a transformation satisfying T^2 = 0.

Not sure where to start? Take the ten-question placement test.