A question involving linear operators $T$ on $V := \text{span}\{1,x,x^2\}$ Let $T:V\mapsto V$ be the linear operator on the quadratics $V := \text{span}\left\{1,x,x^2\right\}$ given by
\begin{equation*}
Tf(x) := f(0)(1-x)^2+f(1)x^2.
\end{equation*}
Find the matrix representation of $T$ with respect to the basis $B = \left[1,x,x^2\right]$ along with the eigenvalues and eigenspaces of $T$ (as quadratic polynomials).
Okay so a vector $v$ of $V$ can be represented as $$v = a\cdot 1+b\cdot x + c x^2 \qquad a,b,c \in \mathbb{R}$$ but how would we apply this to $Tf(x)$?
Thanks.
 A: If I am understanding your question correctly you are attempting to represent your linear operator as a matrix. If that is the case then we know that every vector in the the vector space $V$ can be written as a linear combination of the basis $\{1,x,x^2\}$, similarly the range of your linear operator can be represented as a linear combination of your basis. Since your operator is linear then we only need be concerned with the transformation of the basis, i.e. any other vector found in $V$ its transformation can be represented as the linear combination of the columns of your matrix representation of the basis transformation. 
Consider: 
$$Tf(x)=T(1) = 1(1-x)^2+1*x^2=f(1)=1-2x+2x^2$$
$$Tf(x)=T(x) = 0(1-x)^2+1*x^2=x^2$$
$$Tf(x)=T(x^2)=0^2(1-x)^2+(1)^2*x^2=x^2$$
Then our matrix representation of the linear operator is as follows with columns representing the coordinates of each respective transformation with respect to the basis: 
\begin{pmatrix}
1 & 0 &0 \\
-2 & 0 &  0 \\
2 &1 &1 
\end{pmatrix}
You can decompose this matrix to get eigenvalues and eigenspaces. 
