Linear transformation using bases

Question

Let T: $M_2 (\mathbb{R}) \to P_2$ be a linear transformation determined by T $ \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} $ = 1 - x² ; T $ \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} $ = 2 + x + x²; T $ \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix} $ = -x + x²; T $ \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix} $ = 1 + 2x.

a) Determine T $ \begin{bmatrix} 1 & 3 \\ -2 & 0 \\ \end{bmatrix} $

b) Find T $ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $, for any a, b, c, d $\in \mathbb{R}$

c) Be $\mathcal B = \begin{Bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 1 & 0 \\ \end{bmatrix}, \begin{bmatrix} 2 & 0 \\ 1 & 1 \\ \end{bmatrix} \end{Bmatrix} $ and $\mathcal C$ = {x², 3+x, 2+x+x²} bases from $M_2 (\mathbb{R})$ and from $P_2(\mathbb{R})$ respectively. Determine the matrix T in relation to $\mathcal B$ and $\mathcal C$

Please, help me with this exercise, because I'm not doing this right

Answer 1

In view of b), by linearity, $$T \left(\begin{array}{cc}a & b\\ c & d\end{array}\right) = a T\left(\begin{array}{cc}1 & 0\\ 0 & 0\end{array}\right) + b T\left(\begin{array}{cc}0 & 1\\ 0 & 0\end{array}\right) + c T\left(\begin{array}{cc}0 & 0\\ 1 & 0\end{array}\right) + d T\left(\begin{array}{cc}0 & 0\\ 0 & 1\end{array}\right)= a(1-x^2) + b(2+x+x^2) + c(-x+x^2) +d (1+2x). $$ Expansion gives a real-valued polynomial of degree at most 2.

Answer 2

In base $\mathcal B$ we have the first value $ \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix} $. Writting this value, using the linear transformation from the exercice, I got: T$\begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix}$ = [(1-x²)+(2+x+x²)] = (3+x). Is that right?

If, this is right, I need to describe this in the bases $\mathcal C$. Lookin at the base, I got \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}. Is that right? Or I'm missing something?

Answer 3

You mention in a comment to another answer that you “tried to use (b) in (c).” That’s certainly a valid way to approach the problem instead of essentially starting over from scratch for (c). You’ve found a (hopefully $3\times4$) matrix $M=[T]_{\mathcal E}^{\mathcal E'}$ that expresses $T$ relative to the standard ordered bases $\mathcal E'$ and $\mathcal E$ of $M_2(\mathbb R)$ and $P_2$, respectively. In order to compute $[T]_{\mathcal C}^{\mathcal B}$ you just need to multiply by the appropriate change-of-basis matrices: $$[T]_{\mathcal B}^{\mathcal C} = [\operatorname{id}]_{\mathcal C}^{\mathcal E}[T]_{\mathcal E}^{\mathcal E'}[\operatorname{id}]_{\mathcal E'}^{\mathcal B}.$$ (Here the superscripts denote the “input” bases and subscripts the “output” bases.) Now, the columns of the change-of-basis matrix that converts from $\mathcal B$ to the standard basis are, as always, the images of the elements of $\mathcal B$ expressed relative to the standard basis—i.e., the coordinate vectors of the elements of $\mathcal B$. Calling this matrix $B$, and similarly for the matrix $C$, that means that you have to compute $C^{-1}MB$.

Whether this is more or less work than computing it directly from the definition of $T$ and the two bases depends on what you’re comfortable with. Inverting a $3\times 3$ matrix ($C$) is tedious, but not very difficult, and in fact that part of the product can be computed via row-reduction of the augmented matrix $[C\mid MB]$: when you’re done, you end up with $[I_3\mid C^{-1}MB]$.