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Given $T: P_2 \rightarrow P_3$ defined by:

$T(at^2 + bt +c) = (a-b+c)t^3 + (-a + 3b - 2c)t^2 +(-a-b)t +(2b-c)$

What is the corresponding Matrix of $T$?

This is what I have: First I rewrite the transformation as follows:

$T_0 = 2b - c$

$T_1 = -a - b$

$T_2 = -a + 3b -2c$

$T_3 = a-b+c$

And I know $T_i = \sum \limits_{j=1}^n \mu_{ji} b_i$ where $b_i$ is the basis vector.

So my matrix $\mu$ is $\pmatrix{0&-1&-1&1\\ 2&-1&3&-1\\ -1&0&-2&1}$

  1. Is this correct?
  2. Is my rewriting of $T$ correct?
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If you're using column vectors, your matrix is the wrong size. It should be $4 \times 3$. –  wj32 Nov 11 '12 at 21:09
    
I think you need to transpose $\mu$, as $P_2$ is a 3 dimensional space, mapping to $P_3$, a 4 dimensional space. –  Daryl Nov 11 '12 at 21:09
    
I think you should consider what your basis vectors are first. Your domain should have 3 and the range 4. –  copper.hat Nov 11 '12 at 21:11

1 Answer 1

up vote 2 down vote accepted

Whenever you are working out a problem such as this, the key fact to remember is always that a linear mapping's action is completely determined by it's action on a basis.

I am going to assume that you are using the standard basis vectors for $P_2$ and $P_3$, if not then it is a good exercise to modify the procedure below for your chosen basis. In this case, let us look at how the mapping acts on the basis vectors $$T(1) = t^3 - 2t^2 - 1$$ $$T(t) = -t^3 + 3t^2 - t + 2$$ $$T(t^2) = t^3 - t^2 - t$$ In terms of the standard coordinates, this is then $$T:\ \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}\mapsto\begin{pmatrix}-1 \\ 0 \\-2 \\ 1\end{pmatrix}$$ $$T:\ \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}\mapsto\begin{pmatrix}2 \\ -1 \\ 3 \\ -1\end{pmatrix}$$ $$T:\ \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}\mapsto\begin{pmatrix}0 \\ -1 \\ -1\\ 1\end{pmatrix}$$ Putting this all together gives you the standard matrix $$[T] = \begin{pmatrix} -1 & 2 & 0 \\ 0 & -1 & -1 \\ -2 & 3 & -1 \\ 1 & -1 & 1\end{pmatrix}$$

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