# Linear transformation and diagonalization

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?

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