Find the matrix of  relative to the basis  So I have a math problem, and I already have the answer to it, but I am so confused on a specific part of it. The problem goes as follows:
Let : 1 → 1 be the linear transformation defined for all polynomials () in 1 by
(()) = ( − 1). Find the matrix of  relative to the basis  = {1, } for 1.
So the matrix which is the answer is
$$
\begin{bmatrix}
1 & -1\\
0 & 1\\
\end{bmatrix}
$$
To get the matrix though, I thought I needed to evaluate T(f(x)) at 1 and x. However, isn't T(1) = f(x-1) = 1-1 = 0. I'm being told that when you evaluate it at 1 you get 1 but I don't understand how. I know it can't be 0 because then the linear combination of the basis vector won't result in a vector of
$$
\begin{bmatrix}
1\\
0\\
\end{bmatrix}
$$
needed to make the final matrix, but I just don't understand how it evaluates to 1 and not 0.
 A: You are confusing functions with evaluations .
$T(1)=\mathbf{1}(x-1)=1$. Here $\mathbf{1}$ is just the identity polynomial.
Now if $f(x)=x$. So ,$f(x-1)=x-1$.
So $T(f(x))=f(x-1)=x-1=-1+x=-1.\mathbf{1}+1.x$
So the matrix is
$$\begin{bmatrix} 1 & -1\\
                 0 & 1 \end{bmatrix}$$.
To help you understand better. Consider the polynomial $p(x)=1$.
Then whatever you input for the polynomial , you will end up with $1$. So $p(x-1)=1$.
Remember you are dealing with polynomials, You are not considering evaluations at some $x=c$. for example if $f(x)=x^{2}+1$. Then $f(x-1)=(x-1)^{2}+1=x^{2}-2x+2$. This is a polynomial in it's own right and hence a vector in $P_{2}(\mathbb{F})$.
A: Let $f(x):=a+bx\in P_{1}(\mathbb{R})$ and $ f(x-1)=a+b(x-1)\in P_{1}(\mathbb{R})$.
Now since $\beta=\{1,x\}$ is a canonical basis for the vector space $P_{1}(\mathbb{R})$ so
$$[T(f(x))]_{\beta}=[a+b(x-1)]_{\beta}=\begin{bmatrix} 1 &-1\\ 0& 1 \end{bmatrix}\begin{bmatrix} a\\ b\end{bmatrix}$$
Hence $$[T]_\beta=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}.$$
