# 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*)

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## 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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