## Important result

Let $$A \in \mathbb{F}^{m \times n}$$ where $$\mathbb{F}$$ is a field. Recall that the rank of $$A$$ is given by the dimension of the column space (or row space) of $$A$$. The nullity of $$A$$, denoted by $$\operatorname{nulllity}(A)$$, is the dimension of the nullspace of $$A$$.

Then $\operatorname{rank}(A) + \operatorname{nullity}(A) = n.$

We now give a proof of this result. Recall that elementary row operations do not affect the row space and the nullspace of $$A$$. Let $$R$$ be a matrix in reduced row-echelon form obtained from $$A$$ via elementary row operations. Then, $$\operatorname{rank}(A) = \operatorname{rank}(R)$$ and $$\operatorname{nullity}(A) = \operatorname{nullity}(R)$$.

Note that the dimension of the row space of $$R$$, call it $$k$$, is equal to the number of leading 1's (i.e. pivots) and the dimension of the nullspace of $$R$$ is given by the number of columns of $$R$$ minus the number of pivots. Hence, $\operatorname{rank}(R) + \operatorname{nullity}(R) = k+ (n-k)=n.$ The result now follows.

## Exercise

1. Find the rank of $$\begin{bmatrix} 1 & 2 & 0 \\ -1 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}.$$

2. Can the nullity of a $$4 \times 5$$ matrix with real entries be 0?

3. Let $$R$$ be a matrix in RREF. Prove that the nullity of $$R$$ is given by the number of non-pivot columns.