The Timpson Lyceum

1. (2 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \\ -1 & 4 \end{bmatrix}$ . Compute $T\!\left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right)$ .

Solution

1. $T\!\left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right) = 2\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + 1\begin{bmatrix} -2 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ 2 \end{bmatrix}.$

2. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix}$ . Find all $\mathbf{x}$ whose image is $\mathbf{b} = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$ .

Solution

2. Solve $A\mathbf{x} = \mathbf{b}$ : $\begin{aligned} x_1 + 3x_2 &= 5 \\ 2x_1 + 5x_2 &= 8. \end{aligned}$ Applying $R_2 - 2R_1$ gives $-x_2 = -2$ , so $x_2 = 2$ and $x_1 = 5 - 6 = -1$ . The unique preimage is $\mathbf{x} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ .

3. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 3 \end{bmatrix}$ . Determine whether $\mathbf{b} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ is in the range of $T$ .

Solution

3. $\mathbf{b}$ is in the range iff $A\mathbf{x} = \mathbf{b}$ is consistent. Reduce:

$\left[\begin{array}{cc|c} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 3 & 1 \end{array}\right] \xrightarrow{R_3 - R_1} \left[\begin{array}{cc|c} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right] \xrightarrow{R_3 - R_2} \left[\begin{array}{cc|c} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \end{array}\right].$

Row 3 reads $0 = -1$ , so the system is inconsistent. Therefore $\mathbf{b}$ is not in the range of $T$ .

4. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ . Describe the geometric effect of $T$ on $\mathbb{R}^2$ , and compute the image of the point $(2, -1)$ .

Solution

4. The columns are $T(\mathbf{e}_1) = (0, 1)$ and $T(\mathbf{e}_2) = (-1, 0)$ : the standard basis vectors are each rotated a quarter turn counterclockwise. So $T$ is rotation about the origin by $90^\circ$ counterclockwise. The image of $(2, -1)$ is

$A\begin{bmatrix} 2 \\ -1 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 1 \end{bmatrix} + (-1)\begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.$

5. (3 pts) Suppose $T : \mathbb{R}^2 \to \mathbb{R}^2$ is linear with $T(\mathbf{u}) = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $T(\mathbf{v}) = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$ . Find $T(3\mathbf{u} - 2\mathbf{v})$ and $T(\mathbf{0})$ .

Solution

5. By linearity, $T(3\mathbf{u} - 2\mathbf{v}) = 3T(\mathbf{u}) - 2T(\mathbf{v}) = 3\begin{bmatrix} 2 \\ 1 \end{bmatrix} - 2\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix} - \begin{bmatrix} -2 \\ 6 \end{bmatrix} = \begin{bmatrix} 8 \\ -3 \end{bmatrix}$ . Also $T(\mathbf{0}) = T(0\cdot\mathbf{u}) = 0\cdot T(\mathbf{u}) = \mathbf{0}$ .

6. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) Every linear transformation $T$ satisfies $T(\mathbf{0}) = \mathbf{0}$ .

(b) The range of a transformation $T : \mathbb{R}^n \to \mathbb{R}^m$ is a subset of its domain.

(c) If $T(\mathbf{x}) = A\mathbf{x}$ with $A$ of size $m \times n$ , then the domain of $T$ is $\mathbb{R}^m$ .

Solution

6.

(a) True. $T(\mathbf{0}) = T(0\cdot\mathbf{v}) = 0\cdot T(\mathbf{v}) = \mathbf{0}$ by the scaling property.

(b) False. The range consists of images $T(\mathbf{x})$ , which lie in the codomain $\mathbb{R}^m$ , not the domain $\mathbb{R}^n$ (unless $m = n$ ).

(c) False. $A$ is $m \times n$ , so it acts on vectors with $n$ entries; the domain is $\mathbb{R}^n$ , and the codomain is $\mathbb{R}^m$ .

7. (3 pts) Let $A$ be an $m \times n$ matrix and define $T(\mathbf{x}) = A\mathbf{x}$ . Prove that $T$ is a linear transformation, i.e. that $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{u}) = cT(\mathbf{u})$ for all $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $c$ .

Solution

7. Let $\mathbf{a}_1, \dots, \mathbf{a}_n$ be the columns of $A$ . By definition $A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ . For $\mathbf{u} = (u_1, \dots, u_n)$ and $\mathbf{v} = (v_1, \dots, v_n)$ ,

$T(\mathbf{u} + \mathbf{v}) = \sum_{j} (u_j + v_j)\mathbf{a}_j = \sum_j u_j\mathbf{a}_j + \sum_j v_j\mathbf{a}_j = A\mathbf{u} + A\mathbf{v} = T(\mathbf{u}) + T(\mathbf{v}),$

and for a scalar $c$ ,

$T(c\mathbf{u}) = \sum_j (cu_j)\mathbf{a}_j = c\sum_j u_j\mathbf{a}_j = c(A\mathbf{u}) = cT(\mathbf{u}).$

Both defining properties hold, so $T$ is linear. $\blacksquare$

8. (4 pts ••) Fix a nonzero vector $\mathbf{b} \in \mathbb{R}^n$ and define the translation $T(\mathbf{x}) = \mathbf{x} + \mathbf{b}$ . Prove that $T$ is not a linear transformation. State the general property of linear maps that translations by $\mathbf{b} \neq \mathbf{0}$ violate.

Solution

8. Every linear transformation must map $\mathbf{0}$ to $\mathbf{0}$ . But $T(\mathbf{0}) = \mathbf{0} + \mathbf{b} = \mathbf{b} \neq \mathbf{0}$ (since $\mathbf{b} \neq \mathbf{0}$ ), so $T$ cannot be linear.

To confirm directly, the additivity property fails: for any $\mathbf{u}, \mathbf{v}$ ,

$T(\mathbf{u} + \mathbf{v}) = \mathbf{u} + \mathbf{v} + \mathbf{b}, \quad\text{while}\quad T(\mathbf{u}) + T(\mathbf{v}) = (\mathbf{u} + \mathbf{b}) + (\mathbf{v} + \mathbf{b}) = \mathbf{u} + \mathbf{v} + 2\mathbf{b}.$

These differ by $\mathbf{b} \neq \mathbf{0}$ , so $T(\mathbf{u} + \mathbf{v}) \neq T(\mathbf{u}) + T(\mathbf{v})$ . The property violated is preservation of the origin (equivalently, additivity/homogeneity): a nonzero translation is affine, not linear. $\blacksquare$