Find a Basis For the Null Space of a Given $2\times 3$ Matrix

Linear Algebra Problems and Solutions

Problem 132

Let
\[A=\begin{bmatrix}
1 & 1 & 0 \\
1 &1 &0
\end{bmatrix}\] be a matrix.

Find a basis of the null space of the matrix $A$.

(Remark: a null space is also called a kernel.)

 
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Solution.

The null space $\calN(A)$ of the matrix $A$ is by definition
\[\calN(A)=\{ \mathbf{x} \in \R^3 \mid A\mathbf{x}=\mathbf{0} \}.\] In other words, the null space consists of all solutions $\mathbf{x}$ of the matrix equation $A\mathbf{x}=\mathbf{0}$.

So we first determine the solutions of $A\mathbf{x}=\mathbf{0}$ by Gauss-Jordan elimination. The augmented matrix is
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right].
\end{align*}
Subtracting $R_1$ from $R_2$, we reduce the augmented matrix to the reduced row echelon form matrix as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
Thus the solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfies $x_1+x_2=0$, or equivalently $x_1=-x_2$.
Substituting the last equality, we see that solutions are of the form
\[\mathbf{x}=\begin{bmatrix}
-x_2 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.\] Therefore the null space is
\begin{align*}
\calN(A)&=\left \{\mathbf{x} \in \R^3 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \text{ for any } x_2, x_3 \in \R \right \}\\[6pt] &= \mathrm{Sp} \left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \, \right\}.
\end{align*}
Thus, the set $\left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \, \right\}$ is a spanning set for the null space $\calN(A)$.


Now, we check that the vectors $\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}$ are linearly independent.
Consider a linear combination
\[a_1\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+a_2\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} =\mathbf{0}.\] This is equivalent to
\[\begin{bmatrix}
-a_1 \\
a_1 \\
a_2
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] Hence we must have $a_1=a_2=0$.
Since the linear combination equation has only the zero solution, the vectors $\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}$ are linearly independent.


Therefore the set $\left\{ \, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \,
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\, \right\}$ is a linearly independent spanning set, thus it is a basis for the null space $\calN(A)$.


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