show that [T]β is a diagonal matrix

Question

$V = P_1(R), T(a + b(x)) = (6a - 6b) + (12a - 11b)x$, and $β = \{3+4x, 2+3x\}$

Show that $[T]β$ is a diagonal matrix

I am totally confused about how to write down the matrix form of this question.

Could anyone please help me here?

Answer 1

Here $T:P_1\to P_1$ is defined by $T(a+b\,x)=6\,(a-b)+(12\,a-11\,b)\,x$ and $\beta$ is the basis $\{p,q\}$ for $P_1$ where $p(x)=3+4\,x$ and $q(x)=2+3\,x$.

Note that $$ \begin{array}{rcrcr} T(p) & = & \color{red}{-2}\,p & + & \color{blue}{0}\,q \\ T(q) & = & \color{green}{0}\,p & + & (\color{purple}{-3})\,q \end{array} $$ This implies $$ [T]_\beta= \begin{bmatrix} \color{red}{-2} & \color{green}{0} \\ \color{blue}{0} & \color{purple}{-3} \end{bmatrix} $$

Answer 2

For a linear transformation $T:V\to V$ and a basis $\beta=(b_1,b_2,\dots)$ of $V$, you can get the $i$th column of the matrix $[T]_\beta$ by calculating the coordinates of $T(b_i)$, taken in the basis $\beta$.

Being diagonal claims $T(b_i)=\lambda_ib_i$ with some $\lambda_i\in\Bbb R$ for each basis element $b_i$.