4.2.2

Let T be a linear operator on $$\mathbb{R}^2 $$ which carries the line y = x to the line y = 3x. Using the standard basis for $$\mathbb{R}^2 $$, with $$ v = \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $$, we have $$Tv = k\begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} $$, where k is a real number. Let A be the matrix of T under the standard basis, $$ A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$, then we have $$Tv = Av = \begin{bmatrix} a+b \\ c+d \\ \end{bmatrix} = \begin{bmatrix} k \\ 3k \\ \end{bmatrix} $$. This implies $$ b = k-a, \ d = 3k - c$$, so the matrix of T is $$\begin{bmatrix} a & k-a \\ c & 3k - c \\ \end{bmatrix} $$.

Conversely, we can verify that linear operators on $$\mathbb{R}^2 $$ whose matrices with respect to the standard basis have the form $$\begin{bmatrix} a & k-a \\ c & 3k - c \\ \end{bmatrix} $$ will take vectors of the form $$x\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $$ to $$xk\begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} $$, and hence send the line y = x to y = 3x. Therefore the linear operators T on $$\mathbb{R}^2 $$ which send the line y = x to the line y = 3x are exactly all those transformations whose matrices with respect to the standard basis have the form $$\begin{bmatrix} a & k-a \\ c & 3k - c \\ \end{bmatrix} $$, $$ k \in \mathbb{R} $$.

(Note that if k = 0, T sends the entire line y = x to the zero vector. If you want to prevent T from behaving in this way, you can interpret the statement "T carries the line y = x to the line y = 3x" more restrictively as "Given any vector w on y = 3x, there exists some v on y = x such that Tv = w", which eliminates the possibility that k = 0. The proof is then exactly the same.)