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Given the transformation $T : \mathbb{P}_3 \rightarrow \mathbb{P}_4$ defined as $T(\mathbf{p}(t)) = (-t)\mathbf{p}(t)$

where $\mathbf{p}(t) = t^3 + 6t - 2$

I am first supposed to find $T$, which should be easy enough and gives: $$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t$$

Now, what I can't do is to find the matrix $M$ relative to the bases $B = \{1, t, t^2, t^3\}$ and $C = \{1, t, t^2, t^3, t^4\}$

Looking in my book and other threads here, It seemed like the appropriate way is to insert every element of $B$ into $T$ and set up a matrix with the coefficients.

$$T(\mathbf{p}(1)) = -(1)^4 - 6(1)^2 + 2(1) = -5$$ $$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t $$ $$T(\mathbf{p}(t^2)) = -(t^2)^4 - 6(t^2)^2 + 2(t^2) = -t^8 - 6t^4 + 2t^2$$ $$T(\mathbf{p}(t^3)) = -(t^3)^4 - 6(t^3)^2 + 2(t^3) = -t^{12} - 6t^6 + 2t^3$$

which gives

\begin{equation} M = \begin{bmatrix} -5 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & -6 & 2 & 0\\ 0 & 0 & 0 & 2\\ 0 & -1 & -6 & 0 \end{bmatrix} \end{equation}

Happy with any helps or hints, thanks

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  • $\begingroup$ Shouldn't you find where $t^n$ goes, and not this concrete polynomial. $\endgroup$
    – nonuser
    Commented May 1, 2018 at 10:23
  • $\begingroup$ A useful property for problems like these is that the columns of a transformation matrix are the images of the basis vectors. $\endgroup$
    – amd
    Commented May 1, 2018 at 20:24

2 Answers 2

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I suspect you already found the answer to this question based on your comment to ChristianF's answer, but just in case, the transformation matrix is

$ M = \begin{bmatrix}0 & 0 & 0 & 0\\-1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & -1\end{bmatrix}$

with respect to the bases $B$ and $C$ you stated.

If you had selected the bases $B^* = \{t^3,t^2,t,1\}$ and $C^* = \{t^4,t^3,t^2,t,1\}$, you would obtain ChristianF's answer for $M$.

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Since $T(t^n)=-t^{n+1}$ i would say it is: \begin{equation} M = \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation} Then \begin{equation} \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 6 \\ -2 \end{bmatrix}= \begin{bmatrix} -1 \\ 0 \\ -6 \\ 2\\ 0 \end{bmatrix} \end{equation}

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  • $\begingroup$ Thanks you got me on the right track but the answer was moving the all your diagonal of -1's down one row $\endgroup$ Commented May 1, 2018 at 13:45
  • $\begingroup$ that can't be, then you would not have $t^4$ $\endgroup$
    – nonuser
    Commented May 1, 2018 at 13:50

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