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I am working on the following linear algebra question and have gotten half the correct result, while the remaining transformation is slightly confusing.

The question is as follows:

There is a transformation with the set of real numbers,

$T1 : P_2 ~-> M_2$

$T1\left( a + bx +cx^2 \right) = \begin{bmatrix} a-2c & a+3b \\ a-b-c & -4b \\ \end{bmatrix}$

$T2 : M_2 ~-> R^2$ via

$T2 \left( \left [ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right ] \right) = (-2a, a + b + c + d)$

The question asks for these two transformations to be applied, T1, then T2.

Attempt at a solution: Let A, B, C be standard bases for $P_2, M_2, R_2$ respectively.

We know [$T_2 T_1]^C_A = [T_2]^C_B [T_1]^B_A$

Calculating $[T_2]^C_B$

Standard Basis for B is the set of 2x2 matrices as follows. \begin{bmatrix}1&0\\0&0\end{bmatrix} \begin{bmatrix}0&1\\0&0\end{bmatrix} \begin{bmatrix}0&0\\1&0\end{bmatrix} \begin{bmatrix}0&0\\0&1\end{bmatrix}

The standard basis for C is the bases formed in $R^2$, so (1,0),(0,1).

I then calculated, for example: $T\begin{bmatrix}1&0\\0&0\end{bmatrix}_C = [(-2, 1)]_C$ So:

$(-2,1) = a(1, 0) + b(0,1)$

$a = -2, b = 1$

Following this for each of the 4 bases matrices results in...

$[T_2]^C_B = \begin{bmatrix}-2&0&0&0\\1&1&1&1\end{bmatrix} $

Could anyone provide help calculating the matrix $[T_1]^B_A$ ? According to the solutions I correctly calculated this first matrix, not sure how to approach the second.

I understand the transition from a matrix better than I do from a polynomial basis to a matrix. Any help would be appreciated- for self study. Thank you!

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  • $\begingroup$ In the T2 transformation, where does the d in a+b+c+d come from? $\endgroup$
    – user137481
    Jun 23, 2017 at 21:42
  • $\begingroup$ @user137481 made a bit of a typo - edited to fix the T2 transformation matrix, should contain a, b, c, and d. Thanks for pointing that out! $\endgroup$
    – rubyquartz
    Jun 23, 2017 at 22:54
  • $\begingroup$ The question as you’ve posed it doesn’t say anything about computing matrices for these transformations. As I understand it, it simply asks you to compute the composition $T_2\circ T_1$. Observe that $T_2$ is defined in terms of the output of $T_1$, so it should be pretty simple to put the two together without going through all of the work of building matrix representations for them. $\endgroup$
    – amd
    Jun 23, 2017 at 22:54
  • $\begingroup$ @amd Thanks for pointing that out- I was following the method given by the textbook, but if you know of a more efficient method and you wouldn't mind explaining that'd be a great help. $\endgroup$
    – rubyquartz
    Jun 23, 2017 at 22:55
  • $\begingroup$ Ah, after your edit, the problem isn’t as trivial to solve as it was originally, but I think it’s still simpler to work it out directly. $\endgroup$
    – amd
    Jun 23, 2017 at 22:59

1 Answer 1

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This is a case for which computing the composition of the two transformations directly seems like a bit less work than converting to matrices. We have $$T_1(a+bx+cx^2)=\begin{bmatrix}a-2c&a+3b\\a-b-c&-4b\end{bmatrix}.$$ Plugging this matrix into the definition of $T_2$ gives $$\begin{align}(T_2\circ T_1)(a+bx+cx^2)&=(-2(a-2c),(a-2c)+(a+3b)+(a-b-c)+(-4b))\\&=(-2a+4c,3a-2b-3c).\end{align}$$

To solve this using transformation matrices, “vectorize” $\mathbb R^{2\times2}$ via the isomorphism $$\phi:\begin{bmatrix}a&b\\c&d\end{bmatrix}\mapsto\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$ and compute the matrices of $\phi\circ T_1$ and $T_2\circ\phi^{-1}$. Recall that the columns of a transformation matrix are the images of the basis, so we can easily write down these two matrices relative to the standard bases by setting each of $a$, $b$, $c$ to $1$ and the others to $0$ in turn. The matrix of the composition $T_2\circ T_1=(T_2\circ\phi^{-1})\circ(\phi\circ T_1)$ is their product: $$\begin{bmatrix} -2&0&0&0 \\ 1&1&1&1\end{bmatrix}\begin{bmatrix} 1&0&-2 \\ 1&3&0 \\ 1&-1&-1 \\ 0&-4&0 \end{bmatrix} = \begin{bmatrix} -2&0&4 \\ 3&-2&-3 \end{bmatrix}.$$ From this matrix we can read that $$T_2\circ T_1: a+bx+cx^2\mapsto(-2a+4c,3a-2b-3c)$$ which agrees with the previous calculation.

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