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Define $\alpha$ as $\alpha(p)(x)=(x-1)(p'(x)+p'(1))$ Let V also be the space of polynomials of degree less than or equal to 2. The basis given is ${{1,x,x^{2}}}$

After after what I hope to have been a correct computation, I got the matrix $\mbox{} \left[\begin{array} 11 & x & x^{2} \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{array} \right]$

Now I can get the characteristic polynomial, to get the eigenvalues and then the nullspaces of each eigenvalue, if the number of distinct eigenvalues is equal to 3, then the matrix is diagonalizable, if not then I can find the Jordan canonical form.

Is this all correct?

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No. The entries in the matrix ought to be numbers, not polynomials. –  Gerry Myerson Sep 27 '12 at 4:49
    
@GerryMyerson, ah, I was worried about that. Could you show me how to do 1 iteration of the basis? –  JackinAstoria Sep 27 '12 at 4:50
    
I don't know what you mean by "iteration" in this context, but I'll write a bit of the solution, if no one beats me to it. –  Gerry Myerson Sep 27 '12 at 5:01
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1 Answer 1

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Calculate $\alpha(0)=0$, $\alpha(x)=2(x-1)=-2+2x$, $\alpha(x^2)=(x-1)(2x+2)=-2+2x^2$, so the matrix is $$\pmatrix{0&-2&-2\cr0&2&0\cr0&0&2\cr}$$

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Thank you, I meant for one element of the basis. I didn't read the formula correctly it seems. –  JackinAstoria Sep 27 '12 at 5:38
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