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Let $P_3(R)$ be the vectorspace of all real polynomials $\le 3$, such that the polynomial $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ and let T be the linear operator on $P_3(R)$ that we get by defining $T$ as $T(p(x))=(x^3+x)p''(x)-2x^2p'(x)$ for all polynomials $p(x)\in P_3(R)$. Now decide a basis for the nullspace and the columnvector-space(Is this the same as $Im\,T$ in english?) for $T$ or show that is only contains the zero-polynomial. Furthermore, find all eigenvalues for $T$, and examine if $T$ is diagonalizable(correct translation?)?

$p'(x)=a_1+2a_2x+3a_3x^2$, $p''(x)=2a_2+6a_3x$

$Attempt:$ $T(a_0+a_1x+a_2x^2+a_3x^3)=(x^3+x)(2a_2+6a_3x)-2x^2(a_1+2a_2x+3a_3x^2) = x(2a_2)+x^2(-2a_1-4a_2+6a_3)+x^3(2a_2)=a_0(0)+a_1(-2x^2)+a_2(x-4x^2+2x^3)+a_3(6x^2)$

For the nullspace, let us find the solutions to $T(a_0+a_1x+a_2x^2+a_3x^3)=0$ Which means that $2a_2=-2a_1-4a_2+6a_3=2a_2=0 \to a_1=a_2=a_3=0$ If this is true then $a_0=0$. And the nullspace only contains the zero-polynomial. Im confused here, is this correct?

Edit: After correcting the computation mistake - I tried to convert to matrix-format, and got that the nullspace is spanned by $[1,3x+x^3]$. If this is correct, I should be able to continue.

For the $Im(a)$(?) we have $a_0(0)+a_1(-2x^2)+a_2(x-4x^2+2x^3)+a_3(6x^2)$ so we get the basis $[-2x^2, x-4x^2+2x^3,6x^2]$. I dont really understand this, and it´s probably wrong.

To find the eigenvalues we need to get this in matrix-form, right? How do we translate it into a matrix? If I could do that it would probably be easier for me to uderstand the basis aswell...

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As it stands, $T$ is not an endomorphism of $P_3(\mathbb R)$ as it may produce degree 4 polynomials. Are you sure your explicit expression for $T$ is correct? Better check it once more. –  Hagen von Eitzen Jan 6 '13 at 13:18
    
Im sry, I dont understand endomorphism. The expression of T is as its written in the questionaire. But I realize now that it does produce degree 4 polynomials, which doesnt lie in our vectorspace. How should I interpret that? Ill look up endomorphism :) –  HansDeKling Jan 6 '13 at 13:26
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You should interprete it in such a way that you should rethink your computation of $T$. (And endomorphism means just that: the image is in the original space) –  Hagen von Eitzen Jan 6 '13 at 13:26
    
Ok Hagen thx, but what happens to the degree 4 polynomials? They just disappear or do I degrade them somehow? –  HansDeKling Jan 6 '13 at 14:34
    
If you check your computation carefully, you will find that $\deg f\le 3$ implies $\deg T(f)\le 3$, so everything is fine. If it were not the case, some parts of the problem statement make no (or at least less) sense. –  Hagen von Eitzen Jan 6 '13 at 15:06
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