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 $ImT$ 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...
