Are these vectors in the Nullspace of the Matrix?

Problem 692

Let $A=\begin{bmatrix} 1 & 0 & 3 & -2 \\ 0 &3 & 1 & 1 \\ 1 & 3 & 4 & -1 \end{bmatrix}$. For each of the following vectors, determine whether the vector is in the nullspace $\calN(A)$.

(a) $\begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

(b) $\begin{bmatrix} -4 \\ -1 \\ 2 \\ 1 \end{bmatrix}$

(c) $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

(d) $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Then, describe the nullspace $\calN(A)$ of the matrix $A$.

Solution.

Recall that a vector $\mathbf{v}$ is in the nullspace $\calN(A)$ if $A\mathbf{v}=\mathbf{0}$.

(a) $\begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

We compute
\begin{align*}
A\begin{bmatrix}
-3 \\
0 \\
1 \\
0
\end{bmatrix}=\begin{bmatrix}
1 & 0 & 3 & -2 \\
0 &3 & 1 & 1 \\
1 & 3 & 4 & -1
\end{bmatrix}\begin{bmatrix}
-3 \\
0 \\
1 \\
0
\end{bmatrix}=\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}\neq \mathbf{0}.
\end{align*}
Hence, the vector $\begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}$ is not in $\calN(A)$.

(b) $\begin{bmatrix} -4 \\ -1 \\ 2 \\ 1 \end{bmatrix}$

Since we have
\begin{align*}
A\begin{bmatrix}
-4 \\
-1 \\
2 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0 & 3 & -2 \\
0 &3 & 1 & 1 \\
1 & 3 & 4 & -1
\end{bmatrix}
\begin{bmatrix}
-4 \\
-1 \\
2 \\
1
\end{bmatrix}
=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix},
\end{align*}
the vector $\begin{bmatrix} -4 \\ -1 \\ 2 \\ 1 \end{bmatrix}$ is in $\calN(A)$.

(c) $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

Since
$A\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix},$ we see that $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ is in $\calN(A)$.

(d) $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Note that the size of the matrix $A$ is $3\times 4$. Since $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is an $3$-dimensional vector, the matrix product $A\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is not defined. So in particular, $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is not in $\calN(A)$.

Describe the nullspace $\calN(A)$ of the matrix $A$.

Now, let us describe the nullspace
$\calN(A)=\{\mathbf{x}\in \R^4 \mid A\mathbf{x}=\mathbf{0}\}.$ To do this, we solve the equation $A\mathbf{x}=\mathbf{0}$.
The augmented matrix is
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
1 & 0 & 3 & -2 &0\\
0 &3 & 1 & 1 &0 \\
1 & 3 & 4 & -1 &0
\end{array} \right] \xrightarrow{R_3-R_1}
\left[\begin{array}{rrrr|r}
1 & 0 & 3 & -2 &0 \\
0 & 3 & 1 & 1 & 0 \\
0 & 3 & 1 & 1 & 0
\end{array}\right] \xrightarrow{R_3-R_2}\\[6pt] \left[\begin{array}{rrrr|r}
1 & 0 & 3 & -2 &0 \\
0 & 3 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right] \xrightarrow{\frac{1}{3} R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 3 & -2 &0 \\
0 & 1 & 1/3 & 1/3 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right].
\end{align*}
Thus, the solution is
\begin{align*}
x_1 &=-3 x_3+2x_4\\
x_2&=-\frac{1}{3} x_3 – \frac{1}{3}x_4,
\end{align*}
and the vector form solution is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=
\begin{bmatrix}
-3 x_3+2x_4 \\
-\frac{1}{3} x_3 – \frac{1}{3}x_4 \\
x_3 \\
x_4
\end{bmatrix}
=x_3\begin{bmatrix}
-3 \\
-1/3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
2 \\
-1/3 \\
0 \\
1
\end{bmatrix}.
\end{align*}
Since the nullspace consists of these solutions, we obtain
\begin{align*}
-3 \\
-1/3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
2 \\
-1/3 \\
0 \\
1
\end{bmatrix} \text{ for any } x_3, x_4 \in \R\right \}\\[6pt] &=\Span\left\{\,\begin{bmatrix}
-3 \\
-1/3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
-1/3 \\
0 \\
1
\end{bmatrix} \,\right\}
\end{align*}

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