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I hope you liked the title.

I have a question that is as follows:

Consider the linear transformation $T: P_3(\mathbb{R}) \to P_3(\mathbb{R})$ given by $$T(f(x))=f(0)+f'(x)+f''(x)$$ Where the prime symbol denotes differentiation

a) Determine the matrix representation of $T$ with respect to the standard ordered basis of $P_3(\mathbb{R})$

Note: Standard basis $\{1,x,x^2,x^3\}$

Is this correct?

$ \begin{pmatrix} 0&0&0&0\\1&0&0&0\\2&2&0&0\\0&6&3&0 \end{pmatrix}$

??

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  • $\begingroup$ Is the standard ordered basis of $P_3(\mathbb{R})$ $(1,x,x^2,x^3)$ or is it $(x^3,x^2,x,1)$? $\endgroup$ – JimmyK4542 Jun 25 '14 at 4:07
  • $\begingroup$ @JimmyK4542 The first one, sorry I should have added that in $\endgroup$ – Katie Jun 25 '14 at 4:08
  • $\begingroup$ What did you get when you computed $T(1)$, $T(x)$, $T(x^2)$, $T(x^3)$? That should determine the columns of the matrix. $\endgroup$ – JimmyK4542 Jun 25 '14 at 4:09
  • $\begingroup$ @JimmyK4542 $0,1,2x+2,3x^2 + 6x$ respectively. I think I may have put them in the rows. Isn't it put $T(x)$ in row $2$, get out $1*1, 0*x,0*x^2$ etc $\endgroup$ – Katie Jun 25 '14 at 4:11
  • $\begingroup$ If $f(x) = 1$, then $f(0) = 1$ not $0$. Also, I believe the coefficients for $T(1)$ determine the first column of the matrix, the coefficients for $T(x)$ determine the second column of the matrix, etc. $\endgroup$ – JimmyK4542 Jun 25 '14 at 4:13
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$T(1)=1=1.1+0.x+0.x^2+0.x^3$

$T(x)=1$

$T(x^2)=0+2x+2=2.1+2.x+0.x^2+0.x^3$

$T(x^3)=0+3x^2+6x=0.1+6.x+3.x^2+0.x^3$

I don't see how you got that matrix when the matrix should be $$\begin{pmatrix} 1&1&2&0\\0&0&2&6\\0&0&0&3\\0&0&0&0 \end{pmatrix}$$

If $f(x) \in Ker(T)$, then $T(f(x))=0$ which would give $f(0)+f'(x)+f''(x)=0$ . Now since any $f(x) \in P_3$ will look like $a_0+a_1x+a_2x^2+a_3x^3$, we will have $(a_0)+(a_1+2a_2x+3a_3x^2)+(2a_2+6a_3x)=0$ which gives $(a_0+a_1+2a_2)+x(2a_2+6a_3)+x^2(3a_3)=0$. Now this is $0$ for all $x$. Hence $3a_3=0$, $2a_2+6a_3=0$ and $a_0+a_1+2a_2=0$. Solvin these together will give you $a_0=-a_1$. Hence $f(x) =a_0(1-x)$.

Note you can also find these equations by solving for $X=(a_0,a_1,a_2,a_3)$ such that $AX=0$, where $A$ is your basis matrix

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  • $\begingroup$ Yep, same matrix but I did it row wise instead of column wise, I get it now. Thank you!(and I did T(1) wrong) $\endgroup$ – Katie Jun 25 '14 at 4:15
  • $\begingroup$ How do I find kernel T here> $\endgroup$ – Katie Jun 25 '14 at 5:03
  • $\begingroup$ Or more specifically the basis for ker(T) $\endgroup$ – Katie Jun 25 '14 at 5:07
  • $\begingroup$ @katie now have a look $\endgroup$ – tattwamasi amrutam Jun 25 '14 at 5:22
  • $\begingroup$ Thank you very much for the detailed response, I understand now! $\endgroup$ – Katie Jun 25 '14 at 5:43

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