Problem Set 6: Rank, Nullity, and Bases

Let \[ A = \begin{bmatrix*}[r] 1 & 2 & 0 & 1 & 3 \\ 2 & 4 & 1 & 3 & 5 \\ 1 & 2 & 1 & 2 & 4 \\ 0 & 0 & 1 & 1 & -1 \end{bmatrix*}. \] You may use the Row Reducer to help you calculate.

1. Find $\rank(A)$ and $\nullity(A)$.

   Solution

   We row reduce $A$ to row echelon form.

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

   There are pivots in columns $1$, $3$, and $5$, so $\rank(A) = 3$. Columns $2$ and $4$ have no pivot, so $x_2$ and $x_4$ are free variables and $\nullity(A) = 2$. As a check, $\rank(A) + \nullity(A) = 3 + 2 = 5$, the number of columns of $A$.

2. Is the equation $Ax = b$ consistent for every vector $b$ in $\RR^4$? Explain.

   Solution

   No. The equation $Ax = b$ is consistent exactly when $b$ is in $\col(A)$, and $\dim \col(A) = \rank(A) = 3$. Since $\dim \RR^4 = 4$, the column space is not all of $\RR^4$, so there are vectors $b$ in $\RR^4$ for which $Ax = b$ has no solution. We can also see this from the echelon form: the last row of the echelon form of $A$ is zero, so for a suitable $b$ the augmented matrix reduces to one with a row $[\,0\ 0\ 0\ 0\ 0 \mid c\,]$ with $c \neq 0$.

3. Let \[ w_1 = \begin{bmatrix*}[r] 1 \\ 0 \\ 1 \\ -1 \\ 0 \end{bmatrix*},\qquad w_2 = \begin{bmatrix*}[r] -1 \\ 1 \\ 1 \\ -1 \\ 0 \end{bmatrix*}. \] Explain why $w_1, w_2$ is a basis of $\Nul(A)$.

   Solution

   We multiply:

   and

   So both $w_1$ and $w_2$ are in $\Nul(A)$. The vectors $w_1$ and $w_2$ are linearly independent, since neither is a multiple of the other (look at the first two entries). By part (a), $\dim \Nul(A) = \nullity(A) = 2$. Two linearly independent vectors in a $2$-dimensional subspace automatically span it, so $w_1, w_2$ is a basis of $\Nul(A)$.

1. $A$ is a $3 \times 7$ matrix, and the solution set of $Ax = 0$ is a plane through the origin. What is $\rank(A)$?

   Solution

   The solution set of $Ax = 0$ is $\Nul(A)$, and a plane through the origin is a $2$-dimensional subspace, so $\nullity(A) = 2$. By the Rank-Nullity Theorem, $\rank(A) + \nullity(A) = 7$, so $\rank(A) = 5$.

2. $A$ is a $6 \times 4$ matrix, and for some vector $b$ the equation $Ax = b$ has exactly one solution. What is $\rank(A)$?

   Solution

   If $v_0$ is a solution of $Ax = b$, then the solutions of $Ax = b$ are $x = v_0 + \Nul(A)$. Since there is only one solution, $\Nul(A) = \{0\}$ and $\nullity(A) = 0$. The Rank-Nullity Theorem implies $\rank(A) = 4 - 0 = 4$.

3. $A$ is a $3 \times 5$ matrix, and $\col(A)$ is a line through the origin. How many free variables does the equation $Ax = 0$ have?

   Solution

   A line through the origin is a $1$-dimensional subspace, so $\rank(A) = \dim \col(A) = 1$. By the Rank-Nullity Theorem, $\nullity(A) = 5 - 1 = 4$. The nullity is the number of free variables, so $Ax = 0$ has $4$ free variables.

4. $A$ is a $3 \times 5$ matrix, explain why the solution set of $Ax = 0$ cannot be a line through the origin.

   Solution

   If the solution set of $Ax = 0$ were a line through the origin, then $\nullity(A) = 1$ and the Rank-Nullity Theorem would give $\rank(A) = 5 - 1 = 4$. But the rank is the number of pivots, and a matrix with $3$ rows has at most $3$ pivots, since each row contains at most one pivot. So no such matrix exists.

In this problem we will play a game called Name That Matrix! In each part you are given a description of a linear transformation $T_A : \RR^2 \to \RR^2$, and your job is to find the matrix $A$.

1. Find the matrix $A$ such that $T_A(e_1) = (3,4)$ and $T_A(e_2) = (-1,0)$.

   Solution

   The columns of $A$ are $T_A(e_1)$ and $T_A(e_2)$, so \[ A = \begin{bmatrix*}[r] 3 & -1 \\ 4 & 0 \end{bmatrix*}. \]

2. Find the matrix $A$ such that $T_A$ maps the face on the left to the face on the right.

   Solution

   The face on the left is the unit square, with corners at the origin, $e_1$, $e_2$, and $e_1 + e_2$. The mouth runs along the bottom edge, from the origin to $e_1$, and the face on the right has its mouth along the edge from the origin to $(2, -1)$. So $T_A(e_1) = (2, -1)$, and the other edge at the origin tells us $T_A(e_2) = (2, 2)$. These are the columns of $A$: \[ A = \begin{bmatrix*}[r] 2 & 2 \\ -1 & 2 \end{bmatrix*}. \] As a check, the remaining corner $e_1 + e_2$ is sent to $(2, -1) + (2, 2) = (4, 1)$, which matches the picture.

3. Find the matrix $A$ such that $T_A$ doubles the length of every vector and rotates it $180^\circ$ around the origin.

   Solution

   Rotating a vector $v$ by $180^\circ$ around the origin gives $-v$, and doubling its length gives $-2v$. So $T_A(e_1) = (-2, 0)$ and $T_A(e_2) = (0, -2)$, and \[ A = \begin{bmatrix*}[r] -2 & 0 \\ 0 & -2 \end{bmatrix*}. \]

4. Find the matrix $A$ such that $T_A(1, 1) = (8, 1)$ and $T_A(1, -1) = (2, -3)$.

   Solution

   We need $T_A(e_1)$ and $T_A(e_2)$, so we write $e_1$ and $e_2$ in terms of the vectors we know about: \[ e_1 = \tfrac{1}{2}(1, 1) + \tfrac{1}{2}(1, -1),\qquad e_2 = \tfrac{1}{2}(1, 1) - \tfrac{1}{2}(1, -1). \] Since $T_A$ is linear,

   \[\begin{aligned} T_A(e_1) &= \tfrac{1}{2} T_A(1, 1) + \tfrac{1}{2} T_A(1, -1) = \tfrac{1}{2}(8, 1) + \tfrac{1}{2}(2, -3) = (5, -1), \\ T_A(e_2) &= \tfrac{1}{2} T_A(1, 1) - \tfrac{1}{2} T_A(1, -1) = \tfrac{1}{2}(8, 1) - \tfrac{1}{2}(2, -3) = (3, 2). \end{aligned}\]

   So \[ A = \begin{bmatrix*}[r] 5 & 3 \\ -1 & 2 \end{bmatrix*}. \] As a check, $A(1, 1) = (5 + 3, -1 + 2) = (8, 1)$ and $A(1, -1) = (5 - 3, -1 - 2) = (2, -3)$.

5. Find the matrix $A$ such that $T_A$ reflects the plane across the line consisting of the points $(3t, 4t)$.

   Solution

   Let $v = (3, 4)$, which lies on the line, and let $w = (-4, 3)$, which is perpendicular to the line. Reflecting across the line fixes $v$ and reverses $w$, so $T_A(v) = v$ and $T_A(w) = -w$. Solving $c_1 v + c_2 w = e_1$ and $c_1 v + c_2 w = e_2$ gives \[ e_1 = \tfrac{3}{25} v - \tfrac{4}{25} w,\qquad e_2 = \tfrac{4}{25} v + \tfrac{3}{25} w. \] Since $T_A$ is linear,

   \[\begin{aligned} T_A(e_1) &= \tfrac{3}{25} T_A(v) - \tfrac{4}{25} T_A(w) = \tfrac{3}{25} v + \tfrac{4}{25} w = \tfrac{1}{25}(9 - 16, 12 + 12) = \tfrac{1}{25}(-7, 24), \\ T_A(e_2) &= \tfrac{4}{25} T_A(v) + \tfrac{3}{25} T_A(w) = \tfrac{4}{25} v - \tfrac{3}{25} w = \tfrac{1}{25}(12 + 12, 16 - 9) = \tfrac{1}{25}(24, 7). \end{aligned}\]

   So \[ A = \frac{1}{25} \begin{bmatrix*}[r] -7 & 24 \\ 24 & 7 \end{bmatrix*}. \] As a check, $A(3, 4) = \frac{1}{25}(-21 + 96, 72 + 28) = (3, 4)$ and $A(-4, 3) = \frac{1}{25}(28 + 72, -96 + 21) = (4, -3) = -w$.

6. Find the matrix $A$ such that $T_A$ maps the face on the left to the face on the right.

   Solution

   This time the face on the left is not the unit square. Its corners are the origin, $u_1 = (-2, 1)$, $u_2 = (1, 2)$, and $u_1 + u_2 = (-1, 3)$, with the mouth along the edge from the origin to $u_1$. On the right, the mouth runs along the edge from the origin to $(-1, 2)$, and the other edge at the origin ends at $(3, -1)$. So $T_A(u_1) = (-1, 2)$ and $T_A(u_2) = (3, -1)$.

   Since $u_1, u_2$ is a basis of $\RR^2$, we can write $e_1$ and $e_2$ in terms of them. Solving $c_1 u_1 + c_2 u_2 = e_1$ and $c_1 u_1 + c_2 u_2 = e_2$ gives \[ e_1 = -\tfrac{2}{5} u_1 + \tfrac{1}{5} u_2,\qquad e_2 = \tfrac{1}{5} u_1 + \tfrac{2}{5} u_2. \] Since $T_A$ is linear,

   \[\begin{aligned} T_A(e_1) &= -\tfrac{2}{5} T_A(u_1) + \tfrac{1}{5} T_A(u_2) = -\tfrac{2}{5}(-1, 2) + \tfrac{1}{5}(3, -1) = \tfrac{1}{5}(5, -5) = (1, -1), \\ T_A(e_2) &= \tfrac{1}{5} T_A(u_1) + \tfrac{2}{5} T_A(u_2) = \tfrac{1}{5}(-1, 2) + \tfrac{2}{5}(3, -1) = \tfrac{1}{5}(5, 0) = (1, 0). \end{aligned}\]

   So \[ A = \begin{bmatrix*}[r] 1 & 1 \\ -1 & 0 \end{bmatrix*}. \] As a check, $A u_1 = (-2 + 1, 2) = (-1, 2)$ and $A u_2 = (1 + 2, -1) = (3, -1)$.

Let $P$ be the plane in $\RR^3$ with equation $x + y + z = 0$. The vectors \[ v_1 = \begin{bmatrix*}[r] 1 \\ -1 \\ 0 \end{bmatrix*},\qquad v_2 = \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*} \] lie in $P$, and the vector $w = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ is perpendicular to $P$. Let $\cB = \{v_1, v_2, w\}$.

1. Explain why $\cB$ is a basis of $\RR^3$.

   Solution

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

   \[ \begin{aligned} &\left[\begin{array}{ccc} 1 & 0 & 1 \\ \llap{$-$}1 & 1 & 1 \\ 0 & \llap{$-$}1 & 1 \end{array}\right] \xrightarrow{R_{2} \to R_{2} + R_{1}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & \llap{$-$}1 & 1 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{array}\right] \end{aligned} \]

   Every column has a pivot, so $v_1, v_2, w$ are linearly independent. Since $\dim \RR^3 = 3$, any three linearly independent vectors in $\RR^3$ automatically span $\RR^3$. So $\cB$ is a basis of $\RR^3$.

2. Let $u = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$. Find $[u]_\cB$.

   Solution

   We solve $c_1 v_1 + c_2 v_2 + c_3 w = u$ by row reducing the augmented matrix.

   \[ \begin{aligned} &\left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ \llap{$-$}1 & 1 & 1 & 2 \\ 0 & \llap{$-$}1 & 1 & 3 \end{array}\right] \xrightarrow{R_{2} \to R_{2} + R_{1}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & \llap{$-$}1 & 1 & 3 \end{array}\right] \xrightarrow{R_{3} \to R_{3} + R_{2}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 3 & 6 \end{array}\right] \\[4pt] &\xrightarrow{R_{3} \to \frac{1}{3}\,R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{array}\right] \xrightarrow{R_{1} \to R_{1} - R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & \llap{$-$}1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{array}\right] \\[4pt] &\xrightarrow{R_{2} \to R_{2} - 2\,R_{3}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & \llap{$-$}1 \\ 0 & 1 & 0 & \llap{$-$}1 \\ 0 & 0 & 1 & 2 \end{array}\right] \end{aligned} \]

   So $c_1 = -1$, $c_2 = -1$, $c_3 = 2$, that is, \[ [u]_\cB = \begin{bmatrix*}[r] -1 \\ -1 \\ 2 \end{bmatrix*}. \] As a check, $-v_1 - v_2 + 2w = (-1 - 0 + 2, 1 - 1 + 2, 0 + 1 + 2) = (1, 2, 3)$.

3. Let $u'$ be the reflection of $u$ across the plane $P$. Note that reflecting in $P$ leaves every vector in $P$ where it is and sends $w$ to $-w$. Find $[u']_\cB$.

   Solution

   By part (b), $u = -v_1 - v_2 + 2w$. Reflecting in $P$ fixes $v_1$ and $v_2$ and reverses $w$, so \[ u' = -v_1 - v_2 - 2w, \qquad\text{that is,}\qquad [u']_\cB = \begin{bmatrix*}[r] -1 \\ -1 \\ -2 \end{bmatrix*}. \]

4. Find $u'$.

   Solution

   We convert back to standard coordinates: \[ u' = -v_1 - v_2 - 2w = -\begin{bmatrix*}[r] 1 \\ -1 \\ 0 \end{bmatrix*} - \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*} - 2\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix*}[r] -3 \\ -2 \\ -1 \end{bmatrix*}. \] As a check, $u$ and $u'$ should differ by a multiple of $w$, and indeed $u - u' = (4, 4, 4) = 4w$.

5. Let $L$ be the line through the origin in the direction of $w$. Rotating $\RR^3$ halfway around $L$ sends $w$ to itself and sends every vector in $P$ to its negative. Find the vector $z$ obtained by rotating $u$ halfway around $L$.

   Solution

   This time the $v_1$ and $v_2$ coordinates change sign while the $w$ coordinate stays the same, so \[ z = v_1 + v_2 + 2w = \begin{bmatrix*}[r] 1 \\ -1 \\ 0 \end{bmatrix*} + \begin{bmatrix*}[r] 0 \\ 1 \\ -1 \end{bmatrix*} + 2\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}. \] Note that $z = -u'$: rotating halfway around $L$ is the same as reflecting in $P$ and then negating.