range and basis of range of linear transformation

Question

Let T: $P_2 \to \mathbb R ^2$ be a linear transformation define by $$ T(a+bx+cx^2) = \begin{pmatrix} a-b \\ b+c \\ \end{pmatrix} $$ I'm trying to figure out the range(T) and its basis. From a search online I found the range is the column space, i.e all possible linear combinations of its column vectors. So

$range(T)$ = {$\vec w$ $\vert$ $T (\vec p)$ = $\vec w$}

\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}
Is the only one I can come up with. Can someone explain the process in finding any others in the basis (if they exist)?

Answer 1

Note that

$$T(a+bx+cx^2) = \begin{pmatrix} a-b \\ b+c \\ \end{pmatrix}= a\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}+ b\begin{pmatrix} -1 \\ 1 \\ \end{pmatrix}+ c\begin{pmatrix} 0\\ 1 \\ \end{pmatrix}$$

then we can conclude that the range is $\mathbb R ^2$.

Answer 2

Hint:

In the standard basis $\{1,x,x^2\}$ of $P_2$, the polynomial $a+bx+cx^2$ has components $(a,b,c)^T$, and your transformation is represented bi the matrix: $$ \begin{pmatrix} 1&-1&0\\ 0&1&1 \end{pmatrix} \begin{pmatrix} a\\b\\c \end{pmatrix} =\begin{pmatrix} a-b\\ b+c \end{pmatrix} $$

can you fin the rank of the matrix and a basis for its range?