Matrix representation of linear transformation over the space of polynomials - deducing onto and one-to-one 
Let $\mathcal P_1=\{bx+c: b,\in\mathbb R\}\deg p=1$ define $T: \mathcal P_1\to \mathcal P_1$ by $T(bx+c):=(2b+c)x-c$
  
  
*
  
*How do you get the elements of  $\mathcal P_1$ to a vector $R^2$ and then write T as a 2x2 matrix M.
  

What I just found on a scratch is I wrote (bx+c)=> Matrix [b,c]=[2b+c,-c]=> [2b,c,0,-c]
Would the 2x2 matrix of M just be [2,1,0,-1]?


  
*Explain why T is a one-to-one and onto. (You may use the matrix M, but make sure I know you understand the meaning of one-to-one and onto)
  

For this question I will need to know if I did Number 1 correct.
This was a question I got stuck on to a test that I submitted 2 weeks ago, but my professor hasn't gotten back to my question about the answer and the test is locked down so I am not able to provide what I got.
If anyone can help and explain it and what answer you got that would be tremendous help to knowing how I did on this part.
 A: One answer is as follows:
Let $\mathcal B$ denote the standard basis $\mathcal B = \{1,x\}$ of $\mathcal P_1$.  Then for a polynomial $p(x) = a + bx$, the coordinate vector of $p$ relative to $\mathcal B$ is given by
$$
[p]_{\mathcal B} = \pmatrix{a\\b} =: (a,b).
$$
So in other words, we write the polynomial $a + bx$ as the vector $(a,b) \in \Bbb R^2$.
One way to find the matrix of a transformation relative to a choice of basis is to see what happens to each basis element.  For instance, to find the second column of the transformation, plug $x$ (the second element of $\mathcal B$) to find that
$$
T(x) = T(1 \cdot x + 0) = (2(1) + 0)x - (0) = 0 \cdot 1 + 2\cdot x.
$$
The second column of the matrix of $T$ relative to $\mathcal B$ is the coordinate vector of $T(x)$, which is $(0,2)$.  So the matrix of $T$ has the form
$$
M = [T]_{\mathcal B} = \pmatrix{?&0\\?&2}.
$$
Relative to the basis I have chosen, the matrix is given by $M = \pmatrix{-1&0\\1&2}$. Relative to the basis $\mathcal B' = \{x,1\}$, we would end up with the matrix $M = \pmatrix{2&1\\0&-1}$. Both of these answers are correct.

We could do part 2 without using the matrix as follows:
$T$ is one to one if the only solution to $T(p(x)) = 0$ is $p(x) = 0x + 0 = 0$. We note that if $p(x) = bx + c$, then 
$$
T(p) = 0 \implies (2b + c)x - c = 0 \implies \begin{cases}
2b + c = 0\\
-c = 0
\end{cases} 
$$
Solving this system of equations from the bottom up tells us that $c = 0$ and $b = 0$, so that we must have $p(x) = 0$.  So, $T$ is one to one.
$T$ is onto if the equation $T(p(x)) = q(x)$ has a solution for every $q$.  Writing out $T(p(x)) = q(x)$ where $p(x) = bx + c$ and $q(x) = mx + k$ gives us the system of equations
$$
\begin{cases}
2b + c = m\\
-c = k.
\end{cases} 
$$
It suffices to note that this system of equations has a solution for any $m$ and $k$.
A: Choose the basis $\beta = \{1, x \}$ for $\mathcal{P}_1$. 
Then, the representation of $p = a + bx$ with respect to the basis is called $[p]_\beta$ $$[p]_\beta = \begin{bmatrix}
a\\
b
\end{bmatrix}$$
So in the $\beta$ coordinates, the action of $T(bx + c) = (2b + c)x - c $ translates to: $$T : \begin{bmatrix}
c\\
b
\end{bmatrix} \mapsto \begin{bmatrix}
-c\\
2b + c
\end{bmatrix} $$ 
The associated matrix $[T]_\beta$ is $$[T]_\beta = \begin{bmatrix}
-1 & 0\\
1 & 2
\end{bmatrix} $$
because $$\begin{bmatrix}
-1 & 0\\
1 & 2
\end{bmatrix} \begin{bmatrix}
c\\
b 
\end{bmatrix} = \begin{bmatrix}
-c \\
2b + c 
\end{bmatrix}$$
If $V, W$ are equal dimensional vector spaces and $T : V \to W$ is a linear transformation, then being onto and one-to-one are equivalent. 
Hence, the transformation $T : \mathcal{P}_1 \to \mathcal{P}_1$ is both one-to-one and onto if the associated matrix $[T]_\beta$ is 


*

*invertible

*has nonzero determinant 

*has null space $ = \{0 \}$
Checking any one condition in the above list would suffice.  
