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Let $T:P_2(\mathbb{R})\to P_2(\mathbb{R})$ be the linear transformation on real polynomials of degree not more than 2 defined by:

$$ T(a_2 x^2 + a_1 x + a_0) = (a_0 + a_1)x^2 + (a_1 + a_2)x + (a_0 + a_2). $$

Find a basis for $P_2(\mathbb{R})$ with respect to which the matrix representation for $T$ is diagonal.

I solved that basis for $P_2(\mathbb{R})_\alpha = \{1, x, x^2\}$ and then

$$ [T]_\alpha = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix} $$

and I've diagonalized it. Eigenvalues are -1, 1, 2, and eigenvectors are:

$$ (1, 1, -2)\; (-1, 1, 0)\; (1, 1, 1) $$

Can we say "basis for $P_2(\mathbb{R})$ w.r.t which the matrix representation for $T$ is diagonal" is $\{1, x, x^2\}$ ?

Or : Do we have to say $[T]_\alpha$ is not diagonal matrix, so $\{1, x, x^2\}$ is not the answer?

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A basis for diagonalisation is the same thing as a basis of eigenvectors. Assuming you computed those correctly, your basis would be $1+x-2x^2,-1+x,1+x+x^2$.

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  • $\begingroup$ [T]α is not diagonal matrix. Can we say basis would be 1+x-2x^2, -1+x, 1+x+x^2 $\endgroup$ – linearalgebra Sep 25 '14 at 11:41
  • $\begingroup$ @linearalgebra Please read my answer. $\endgroup$ – Marc van Leeuwen Sep 25 '14 at 12:11

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