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.
Find the rank of \(\begin{bmatrix} 1 & 2 & 0 \\ -1 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}.\)
Can the nullity of a \(4 \times 5\) matrix with real entries be 0?
Let \(R\) be a matrix in RREF. Prove that the nullity of \(R\) is given by the number of non-pivot columns.