Quiz 6. Determine Vectors in Null Space, Range / Find a Basis of Null Space

Introduction to Linear Algebra at the Ohio State University quiz problems and solutions

Problem 313

(a) Let $A=\begin{bmatrix}
1 & 2 & 1 \\
3 &6 &4
\end{bmatrix}$ and let
\[\mathbf{a}=\begin{bmatrix}
-3 \\
1 \\
1
\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}, \qquad \mathbf{c}=\begin{bmatrix}
1 \\
1
\end{bmatrix}.\] For each of the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, determine whether the vector is in the null space $\calN(A)$. Do the same for the range $\calR(A)$.

(b) Find a basis of the null space of the matrix $B=\begin{bmatrix}
1 & 1 & 2 \\
-2 &-2 &-4
\end{bmatrix}$.

 
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Solution of (a)

Let us first recall the definitions of the null space and range. The null space of the $2 \times 3$ matrix $A$ is defined by
\[\calN(A)=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\}\] and the range of $A$ is defined by
\[\calR(A)=\{\mathbf{y} \in \R^2 \mid A\mathbf{x}=\mathbf{y} \text{ for some } \mathbf{x}\in \R^3\}.\] Note that the null space is a subspace of $\R^3$ and the range is a subspace of $\R^2$.

Consider first the vector $\mathbf{a}=\begin{bmatrix}
-3 \\
1 \\
1
\end{bmatrix}$. Since the vector $\mathbf{a}$ is three-dimensional, it cannot be in the range.
Also, since we have
\[A\mathbf{a}=\begin{bmatrix}
1 & 2 & 1 \\
3 &6 &4
\end{bmatrix}
\begin{bmatrix}
-3 \\
1 \\
1
\end{bmatrix}=\begin{bmatrix}
0 \\
1
\end{bmatrix}\neq \mathbf{0},\] the vector $\mathbf{a}$ is not in the null space of $A$.


Next, we consider the vector $\mathbf{b}=\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}$. Again, this is three-dimensional vector, hence $\mathbf{b}$ is not in the range $\calR(A)$.
Since we have
\[A\mathbf{b}=\begin{bmatrix}
1 & 2 & 1 \\
3 &6 &4
\end{bmatrix}
\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}
=\begin{bmatrix}
0 \\
0
\end{bmatrix}=\mathbf{0},\] we see that the vector $\mathbf{b}$ is in the null space $\calN(A)$.


Finally, we consider the vector $\mathbf{c}=\begin{bmatrix}
1 \\
1
\end{bmatrix}$. Since it is two-dimensional, the vector $\mathbf{c}$ is not in the null space $\calN(A)$.
Let us determine whether $\mathbf{c}$ is in the range $\calR(A)$. To do this, we check whether the system $A\mathbf{x}=\mathbf{c}$ is consistent or not. The augmented matrix of the system is
\[[A\mid \mathbf{c}]= \left[\begin{array}{rrr|r}
1 & 2 & 1 & 1 \\
3 & 6 & 4 & 1
\end{array} \right].\] Applying the elementary row operations, we have
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 1 \\
3 & 6 & 4 & 1
\end{array} \right] \xrightarrow{R_2-3R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 1 \\
0 & 0 & 1 & -2
\end{array} \right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 0 & 3 \\
0 & 0 & 1 & -2
\end{array} \right] \end{align*}
and thus it has the general solution
\begin{align*}
x_1&=-2x_2+3\\
x_3&=-2.
\end{align*}
Therefore, the vector $\mathbf{c}$ is in the range $\calR(A)$.
(For example, if $\mathbf{x}_0=\begin{bmatrix}
1 \\
1 \\
-2
\end{bmatrix}$, then $A\mathbf{x}_0=\mathbf{c}$.)

In summary, the vector $\mathbf{a}$ is neither in $\calN(A)$ or in $\calR(A)$. The vector $\mathbf{b}$ is in $\calN(A)$ but not in $\calR(A)$. The vector $\mathbf{c}$ is not in $\calN(A)$ but in $\calR(A)$.

Solution of (b)

We first describe the null space $\calN(B)$ of the matrix $B$.
The vector $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ is in the null space $\calN(B)$ if and only if $B\mathbf{x}=\mathbf{0}$.
The augmented matrix of $B\mathbf{x}=\mathbf{0}$ is
\[ \left[\begin{array}{rrr|r}
1 & 1 & 2 & 0 \\
-2 & -2 & -4 & 0
\end{array} \right].\] Applying the elementary row operations, we have
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 2 & 0 \\
-2 & -2 & -4 & 0
\end{array} \right] \xrightarrow{R_2+2R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right] \end{align*}
and the general solution is
\[x_1=-x_2-2x_3.\] Therefore, the vector $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ is in the null space $\calN(B)$ if and only if $x_1=-x_2-2x_3$.
Thus,
\begin{align*}
\calN(B)&=\{\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\in \R^3 \mid x_1=-x_2-2x_3\}\\
&=\{\mathbf{x}\in \R^3 \mid \mathbf{x}=\begin{bmatrix}
-x_2-2x_3 \\
x_2 \\
x_3
\end{bmatrix} \}\\
&=\{\mathbf{x}\in \R^3 \mid \mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}\}\\
&=\Span \left\{\, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix} \,\right\}.
\end{align*}
Thus, the set
\[S=\left\{\, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix} \, \right\}\] is a spanning set of the null space $\calN(B)$.
Let us verify that $S$ is a linearly independent set. Suppose that we have
\[c_1\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+c_2\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=\mathbf{0},\] for some scalars $c_1, c_2$. Then we have
\[\begin{bmatrix}
-c_1-2c_2 \\
c_1 \\
c_2
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix},\] and thus $c_1=c_2=0$ and $S$ is linearly independent.
Hence $S$ is a linearly independent spanning set, and thus $S$ is a basis of the null space $\calN(B)$.

Comment.

These are Quiz 6 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.


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