# Given a Spanning Set of the Null Space of a Matrix, Find the Rank

## Problem 303

Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \text{ and } \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}.$ Then find the rank of the matrix $A$.

(Purdue University, Linear Algebra Final Exam Problem)

## Solution.

We first determine the nullity of $A$ and deduce the rank of $A$ by the rank-nullity theorem.
The null space $\calN(A)$ of the matrix $A$ is spanned by
$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \text{ and } \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}.$

Let us find a basis of the null space $\calN(A)$ among these vectors.
We use the “leading 1 method”.
Form the matrix whose column vectors are these three vectors and we apply elementary row operations as follows.
\begin{align*}
\begin{bmatrix}
1 & 2 & 1 \\
2 &1 &-1 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & 2 & 1 \\
0 &-3 & -3 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-\frac{1}{3}R_2}\10pt] \begin{bmatrix} 1 & 2 & 1 \\ 0 &1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1-2R_2} \begin{bmatrix} 1 & 0 & -1 \\ 0 &1 & 1 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*} The last matrix is in reduced row echelon form and the first and second column contain the leading 1’s. Therefore, the first two vectors \[\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} form a basis of the null space $\calN(A)$.
Hence the nullity, which is the dimension of $\calN(A)$, is $2$.

Since the size of the matrix $A$ is $7 \times 3$, the rank-nullity theorem gives
$3=\text{nullity of } A + \text{rank of } A.$ Thus, the rank of $A$ is $1$.

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