The Timpson Lyceum

1. (2 pts) Find the standard matrix of the linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^3$ given by $T(x_1, x_2) = (x_1 + 2x_2,\; 3x_1 - x_2,\; x_2)$ .

Solution

1. The columns of the standard matrix are $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ . Here $T(\mathbf{e}_1) = T(1,0) = (1, 3, 0)$ and $T(\mathbf{e}_2) = T(0,1) = (2, -1, 1)$ . Thus

$A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \\ 0 & 1 \end{bmatrix}.$

2. (3 pts) Find the standard matrix of the linear transformation that reflects each point of $\mathbb{R}^2$ across the line $y = x$ .

Solution

2. Reflection across $y = x$ sends $(x_1, x_2) \mapsto (x_2, x_1)$ , swapping coordinates. So $T(\mathbf{e}_1) = T(1,0) = (0,1)$ and $T(\mathbf{e}_2) = T(0,1) = (1,0)$ , giving

$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$

3. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}$ . Determine whether $T$ is onto $\mathbb{R}^2$ and whether $T$ is one-to-one. Justify using pivots.

Solution

3. $A$ is already in echelon form with pivots in columns 1 and 2, i.e. a pivot in each of the two rows, so the columns span $\mathbb{R}^2$ : $T$ is onto. There are three columns but only two pivots, so column 3 has no pivot and $A\mathbf{x} = \mathbf{0}$ has a free variable (a nontrivial solution): $T$ is not one-to-one.

4. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \\ 1 & 1 \end{bmatrix}$ . Determine whether $T$ is one-to-one and whether $T$ is onto $\mathbb{R}^3$ . Justify using pivots.

Solution

4. Reduce $A$ :

$\begin{bmatrix} 1 & 2 \\ 2 & 5 \\ 1 & 1 \end{bmatrix} \xrightarrow[R_3 - R_1]{R_2 - 2R_1} \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 0 & -1 \end{bmatrix} \xrightarrow{R_3 + R_2} \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}.$

Both columns have pivots, so the columns are linearly independent: $T$ is one-to-one. But there are only two pivots among three rows, so the columns cannot span $\mathbb{R}^3$ : $T$ is not onto.

5. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) A linear transformation $T$ is one-to-one if and only if the columns of its standard matrix are linearly independent.

(b) A linear transformation $T : \mathbb{R}^5 \to \mathbb{R}^3$ can be one-to-one.

(c) A linear transformation $T : \mathbb{R}^3 \to \mathbb{R}^5$ can be onto.

Solution

5.

(a) True. $T$ is one-to-one iff $A\mathbf{x} = \mathbf{0}$ has only the trivial solution iff the columns of $A$ are linearly independent (Theorem 12).

(b) False. The standard matrix is $3 \times 5$ ; its five columns are five vectors in $\mathbb{R}^3$ , which must be linearly dependent (more vectors than entries). So $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution and $T$ is not one-to-one.

(c) False. The standard matrix is $5 \times 3$ ; three columns can produce at most three pivots, so they cannot span the five-dimensional $\mathbb{R}^5$ . Hence $T$ cannot be onto.

6. (3 pts) Suppose $T : \mathbb{R}^4 \to \mathbb{R}^3$ is linear and onto. Explain what this forces about the number of pivots in its standard matrix, and use that to decide whether $T$ can be one-to-one.

Solution

6. If $T$ is onto, its standard matrix $A$ ( $3 \times 4$ ) must have a pivot in every one of its $3$ rows, so $A$ has exactly $3$ pivots. Since there are $4$ columns, one column has no pivot, giving a free variable in $A\mathbf{x} = \mathbf{0}$ — a nontrivial solution. Therefore $T$ is not one-to-one. (A map from a higher-dimensional space onto a lower-dimensional one cannot be injective.)

7. (3 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ . Prove that $T$ is one-to-one if and only if the equation $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

Solution

7. ( $\Rightarrow$ ) Suppose $T$ is one-to-one. Since $T$ is linear, $T(\mathbf{0}) = \mathbf{0}$ . If $A\mathbf{x} = \mathbf{0}$ had a solution $\mathbf{x} \neq \mathbf{0}$ , then $T(\mathbf{x}) = \mathbf{0} = T(\mathbf{0})$ with $\mathbf{x} \neq \mathbf{0}$ , contradicting injectivity. So $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

( $\Leftarrow$ ) Suppose $A\mathbf{x} = \mathbf{0}$ has only the trivial solution. If $T(\mathbf{u}) = T(\mathbf{v})$ , then by linearity $A(\mathbf{u} - \mathbf{v}) = A\mathbf{u} - A\mathbf{v} = \mathbf{0}$ , so $\mathbf{u} - \mathbf{v} = \mathbf{0}$ , i.e. $\mathbf{u} = \mathbf{v}$ . Hence $T$ is one-to-one. $\blacksquare$

8. (4 pts ••) Prove: if $T : \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation whose standard matrix is $A$ , then

(a) $T$ onto implies $n \ge m$ ;

(b) $T$ one-to-one implies $n \le m$ .

Solution

8. Let $A$ be the $m \times n$ standard matrix of $T$ . The number of pivots of $A$ is at most $\min(m, n)$ .

(a) If $T$ is onto, the columns of $A$ span $\mathbb{R}^m$ , which requires a pivot in every row, so $A$ has $m$ pivots. Since the number of pivots is at most $n$ , we get $m \le n$ , i.e. $n \ge m$ .

(b) If $T$ is one-to-one, the columns are linearly independent, which requires a pivot in every column, so $A$ has $n$ pivots. Since the number of pivots is at most $m$ , we get $n \le m$ . $\blacksquare$