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}$


  • $\begingroup$ 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





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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.