Linear transformation satisfying $T(x+x^2) = 2x$, $T(1 + x) = x^2$ and $T(1) = 1$.

Question

Let $T : P_2(\mathbb{R}) \to P_2(\mathbb{R})$ be a linear transformation satisfying $T(x+x^2) = 2x$, $T(1 + x) = x^2$ and $T(1) = 1$. Find $T(2x^2 + x)$.

I don't know if I'm doing this right. $T(x^2+x) = 2x + T(x^2) =$ ? which will get me my answer if I add up both transformations?

Answer 1

The key here is to first write $2x^2+x$ as a linear combination of $1$, $1+x$, and $x+x^2$ (one can check that these three polynomials form a basis for $P_2(\Bbb{R})$, so it is possible), and then you appeal to linearity of $T$ to find the image.

Specifically, $$2x^2+x = 2(x^2 + x) - x = 2(x^2+x) - (x + 1) + 1$$

Now from linearity of $T$, we have $$T(2x^2+x) = 2 T(x^2+x) - T(x+1) + T(1) = 2(2x) - x^2 + 1 = 1 + 4x - x^2$$

Answer 2

Let $$ M = \begin{bmatrix} 0 & 0 & 1\\ 2 & 0 & 0\\ 0 & 1 & 0\\ \end{bmatrix} $$

$M$ is the matrix of $T$ written in bases $(x+x^2, 1+x, 1)$ and $(1, x, x^2)$. To get the matrix of $T$ in the standard basis, we can first observe, that $$ B = \begin{bmatrix} 0 & 1 & 1\\ 1 & 1 & 0\\ 1 & 0 & 0\\ \end{bmatrix} $$ transforms a vector written in the first of the bases in to the canonical basis. Therefore, the inverse operation will be the inverse of $B$, $B^{-1}$. After computing: $$ B^{-1} = \begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & -1\\ 1 & -1 & 1\\ \end{bmatrix} $$ Therefore, the matrix of $T$ in standard basis is $$ M\cdot B^{-1} = \begin{bmatrix} 1 & -1 & 1\\ 0 & 0 & 2\\ 0 & 1 & -1\\ \end{bmatrix} $$

To get the result, just multiply it by the vector: $$ M\cdot B^{-1} \cdot [0, 1, 2]^T = [1, 4, -1]^T $$ or, in other words, $1 + 4x - x^2$.

While the above answer details what goes on in the solution (and is probably more valuable), this one provides tools to actually solve problems effectively.