Problem Set 5: Subspaces, Bases, and Dimension

Let \[ A = \begin{bmatrix*}[r] 1 & 0 & 2 & -1 \\ 1 & 1 & 1 & 2 \\ 2 & -1 & 5 & -5 \end{bmatrix*}. \] You may use the Row Reducer to help you calculate.

1. Find a basis for $\Nul(A)$. What is $\dim \Nul(A)$?

   Solution

   $\Nul(A)$ is the set of solutions of the homogeneous system $Ax = 0$, so we row reduce $A$.

   \[ \begin{aligned} &\left[\begin{array}{cccc} 1 & 0 & 2 & \llap{$-$}1 \\ 1 & 1 & 1 & 2 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{cccc} 1 & 0 & 2 & \llap{$-$}1 \\ 0 & 1 & \llap{$-$}1 & 3 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{cccc} 1 & 0 & 2 & \llap{$-$}1 \\ 0 & 1 & \llap{$-$}1 & 3 \\ 0 & \llap{$-$}1 & 1 & \llap{$-$}3 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cccc} 1 & 0 & 2 & \llap{$-$}1 \\ 0 & 1 & \llap{$-$}1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{aligned} \] Columns $3$ and $4$ have no pivots, so $x_3$ and $x_4$ are free, and the reduced matrix says $x_1 = -2x_3 + x_4$ and $x_2 = x_3 - 3x_4$. The general solution in parametric vector form is \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix*}[r] -2 \\ 1 \\ 1 \\ 0 \end{bmatrix*} + x_4 \begin{bmatrix*}[r] 1 \\ -3 \\ 0 \\ 1 \end{bmatrix*}. \] So $\Nul(A) = \sp\{v_1, v_2\}$ where \[ v_1 = \begin{bmatrix*}[r] -2 \\ 1 \\ 1 \\ 0 \end{bmatrix*},\qquad v_2 = \begin{bmatrix*}[r] 1 \\ -3 \\ 0 \\ 1 \end{bmatrix*}. \] These two vectors are linearly independent, since neither is a multiple of the other (look at the last two entries), so $v_1, v_2$ is a basis for $\Nul(A)$ and $\dim \Nul(A) = 2$.

2. Find a basis for $\col(A)$. What is $\dim \col(A)$?

   Solution

   The pivot columns of $A$ form a basis for the column space $\col(A)$. From the row reduction in part (a), the pivots are in columns $1$ and $2$, so a basis for $\col(A)$ consists of the first two columns of $A$, \[ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix},\qquad \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*}, \] and thus $\dim \col(A) = 2$.

3. Is the vector $b = \begin{bmatrix*}[r] 2 \\ -1 \\ 7 \end{bmatrix*}$ in $\col(A)$?

   Solution

   The vector $b$ belongs to $\col(A)$ exactly when the system $Ax = b$ has a solution, so we row reduce the augmented matrix.

   \[ \begin{aligned} &\left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & 2 \\ 1 & 1 & 1 & 2 & \llap{$-$}1 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 & 7 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & 2 \\ 0 & 1 & \llap{$-$}1 & 3 & \llap{$-$}3 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 & 7 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & 2 \\ 0 & 1 & \llap{$-$}1 & 3 & \llap{$-$}3 \\ 0 & \llap{$-$}1 & 1 & \llap{$-$}3 & 3 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & 2 \\ 0 & 1 & \llap{$-$}1 & 3 & \llap{$-$}3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{aligned} \]

   There is no pivot in the augmented column, so the system is consistent and thus $b$ does belong to $\col(A)$. Taking $x_3 = x_4 = 0$ gives the solution $x_1 = 2$, $x_2 = -3$, so \[ b = 2 \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} - 3 \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*}. \]

4. Is the vector $v = \begin{bmatrix*}[r] -3 \\ 4 \\ 1 \\ -1 \end{bmatrix*}$ in $\Nul(A)$?

   Solution

   The vector $v$ is in $\Nul(A)$ exactly when $Av = 0$, and \[ Av = \begin{bmatrix*}[r] 1 & 0 & 2 & -1 \\ 1 & 1 & 1 & 2 \\ 2 & -1 & 5 & -5 \end{bmatrix*} \begin{bmatrix*}[r] -3 \\ 4 \\ 1 \\ -1 \end{bmatrix*} = \begin{bmatrix} -3 + 0 + 2 + 1 \\ -3 + 4 + 1 - 2 \\ -6 - 4 + 5 + 5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}. \] So $v$ is in $\Nul(A)$.

5. (Challenge) Find an equation in $b_1, b_2, b_3$ that is satisfied exactly when the vector $b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$ belongs to the column space $\col(A)$.

   Solution

   We row reduce the augmented matrix $[\,A \mid b\,]$ with a general right hand side, following the same steps as in part (c). \[ \begin{aligned} &\left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & b_1 \\ 1 & 1 & 1 & 2 & b_2 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 & b_3 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & b_1 \\ 0 & 1 & \llap{$-$}1 & 3 & b_2 - b_1 \\ 2 & \llap{$-$}1 & 5 & \llap{$-$}5 & b_3 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - 2\,R_{1}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & b_1 \\ 0 & 1 & \llap{$-$}1 & 3 & b_2 - b_1 \\ 0 & \llap{$-$}1 & 1 & \llap{$-$}3 & b_3 - 2b_1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cccc|c} 1 & 0 & 2 & \llap{$-$}1 & b_1 \\ 0 & 1 & \llap{$-$}1 & 3 & b_2 - b_1 \\ 0 & 0 & 0 & 0 & b_2 + b_3 - 3b_1 \end{array}\right] \end{aligned} \] The last row is the equation $0 = b_2 + b_3 - 3b_1$. The system is consistent exactly when this holds, so $b$ is in $\col(A)$ if and only if \[ b_2 + b_3 - 3b_1 = 0. \]

For each of the following collections of vectors, decide whether it is a basis of $\RR^3$. Explain your answer.

1. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix},\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\quad v_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} } \)

   Solution

   We row reduce the matrix whose columns are $v_1, v_2, v_3$.

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - 2\,R_{1}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}2 \\ 0 & 1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}2 \\ 0 & 0 & 3 \end{array}\right] \xrightarrow{R_{3} \to \frac{1}{3}\,R_{3}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \llap{$-$}2 \\ 0 & 0 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{2} \to R_{2} + 2\,R_{3}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{1} \to R_{1} - R_{3}} \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} \] Every column has a pivot, so $v_1, v_2, v_3$ are linearly independent, and every row has a pivot, so they span $\RR^3$. This collection is a basis of $\RR^3$.

2. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix},\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\quad v_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\quad v_4 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} } \)

   Solution

   This collection is not a basis of $\RR^3$. Four vectors in $\RR^3$ are always linearly dependent: the $3 \times 4$ matrix with these columns has at most $3$ pivots, so at least one column has no pivot and the homogeneous system has a nontrivial solution. In fact, by part (a) the vectors $v_1, v_2, v_3$ already span $\RR^3$, so $v_4$ is a linear combination of them.

3. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix},\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} } \)

   Solution

   This collection is not a basis of $\RR^3$. Two vectors cannot span $\RR^3$: the $3 \times 2$ matrix with these columns has at most $2$ pivots, so some row has no pivot, and the system $Ax = b$ is inconsistent for some $b$ in $\RR^3$. Geometrically, $\sp\{v_1, v_2\}$ is a plane through the origin, and vectors off that plane are not in the span.

4. \( \blue{ v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\quad v_3 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} } \)

   Solution

   We row reduce the matrix whose columns are $v_1, v_2, v_3$.

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] The third column has no pivot, so the vectors are linearly dependent. The reduced matrix says $v_3 = v_1 + v_2$, which we can also see directly. This collection is not a basis of $\RR^3$.

The vectors \[ v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1 \end{bmatrix},\qquad v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} \] are linearly independent.

1. Explain why $v_1, v_2$ is not a basis of $\RR^4$.

   Solution

   Every basis of $\RR^4$ has exactly $4$ vectors, since $\dim \RR^4 = 4$. Concretely, the $4 \times 2$ matrix with columns $v_1, v_2$ has at most $2$ pivots, so two of its rows have no pivot, and $x_1 v_1 + x_2 v_2 = b$ is inconsistent for some $b$ in $\RR^4$. So $v_1, v_2$ do not span $\RR^4$.

2. Find a basis of $\RR^4$ that includes $v_1$ and $v_2$.

   Solution

   We need two more vectors, and there are many choices. Let's try adding \[ v_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix},\qquad v_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \] and row reduce the matrix whose columns are $v_1, v_2, v_3, v_4$.

   \[ \begin{aligned} &\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - R_{1}} \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}\right] \\[4pt] &\xrightarrow{R_{4} \to R_{4} - R_{1}} \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} - R_{2}} \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \end{aligned} \] Every column has a pivot, so $v_1, v_2, v_3, v_4$ are linearly independent, and every row has a pivot, so they span $\RR^4$. So $v_1, v_2, v_3, v_4$ is a basis of $\RR^4$ that includes $v_1$ and $v_2$.

Let $V = \sp\{v_1, v_2\}$, where \[ v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\qquad v_2 = \begin{bmatrix*}[r] 1 \\ -1 \\ 2 \end{bmatrix*}, \] and let $\cB = \{v_1, v_2\}$.

1. Explain why $\cB$ is a basis of $V$.

   Solution

   The vectors $v_1, v_2$ span $V$ by the definition of $V$. They are linearly independent because neither is a multiple of the other. A linearly independent collection that spans $V$ is a basis of $V$, so $\cB$ is a basis of $V$, and $\dim V = 2$.

2. The vector $w = \begin{bmatrix*}[r] 2 \\ 4 \\ -2 \end{bmatrix*}$ is in $V$. Find $[w]_\cB$.

   Solution

   We need to solve $c_1 v_1 + c_2 v_2 = w$, which is a system of three equations in the two unknowns $c_1, c_2$.

   \[ \begin{aligned} &\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 1 & \llap{$-$}1 & 4 \\ 0 & 2 & \llap{$-$}2 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & \llap{$-$}2 & 2 \\ 0 & 2 & \llap{$-$}2 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & \llap{$-$}2 & 2 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{2} \to -\frac{1}{2}\,R_{2}} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 0 \end{array}\right] \\[4pt] &\xrightarrow{R_{1} \to R_{1} - R_{2}} \left[\begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & \llap{$-$}1 \\ 0 & 0 & 0 \end{array}\right] \end{aligned} \] So $c_1 = 3$ and $c_2 = -1$, that is, \[ [w]_\cB = \begin{bmatrix*}[r] 3 \\ -1 \end{bmatrix*}. \] As a check, $3v_1 - v_2 = (3 - 1, 3 + 1, 0 - 2) = (2, 4, -2)$.

3. Is the vector $u = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ in $V$? If it is, find $[u]_\cB$.

   Solution

   We try to solve $c_1 v_1 + c_2 v_2 = u$.

   \[ \begin{aligned} &\left[\begin{array}{cc|c} 1 & 1 & 1 \\ 1 & \llap{$-$}1 & 2 \\ 0 & 2 & 3 \end{array}\right] \xrightarrow{R_{2} \to R_{2} - R_{1}} \left[\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & \llap{$-$}2 & 1 \\ 0 & 2 & 3 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & \llap{$-$}2 & 1 \\ 0 & 0 & 4 \end{array}\right] \end{aligned} \] The last row is the equation $0 = 4$, so the system is inconsistent. There is no way to write $u$ as a linear combination of $v_1$ and $v_2$, so $u$ is not in $V$, and it has no $\cB$-coordinates.