Find formula for linear transformation given matrix and bases

Let T: $$\mathbb P_2\to \mathbb R^3$$ be the linear transformation with matrix $$[T]_{B,A}=\begin{bmatrix}1&1&-1\cr 0&-1&-1\cr -1&0&1 \end{bmatrix}$$ relative to the bases $$A = \{1,2-3x.1+x^2\}$$ and $$B = \{(1,1,1),(1,1,0),(1,0,0)\}$$ find the formula for the linear transformation T.

I don't know what the [T] is supposed to stand for and how to use that information. I'm pretty confident that the linear transformation is supposed to map from B to A and we just need to find the formula to do so.

I know that a similar question has been asked on Finding Linear Transformation with bases and matrix but it didnt explain the steps at all

$$[T]_{B,A}$$ means that if you multiply by the vector of coefficients on $$A$$ you get vector of coefficients of $$B$$. Reciprocally, notice that applying a linear transformation to the basis $$A$$ will give you some vector in the span of the basis $$B$$ the columns vectors of the matrix $$[T]$$ carry the information of how to get the vectors as linear combination of the basis $$B.$$
For example, $$1=(1,0,0)$$ in $$A$$ and if you multiply by $$[T]$$ you get $$(1,0,-1)$$ which corresponds to $$(1,1,1)+0\cdot (1,1,0)-(1,0,0)=(0,1,1)$$ and so $$T(1)=(0,1,1).$$ Similarly $$T(2-3x)=(0,0,1).$$
Can you get $$T(1+x^2)$$?
• I multiplied $[T]$ by $(0,1,0).$ You are multiplying by the basis $\{1,x,x^2\}$ but the basis here is $\{1,2-3x,1+x^2\}$ Jul 19 '20 at 23:55
• Notice that your this vectors come from the representation of each vector in the basis. In the basis $A$ $1+x^2=\color{red}{0}\cdot 1+\color{red}{0}\cdot (2-3x)+\color{red}{1}(1+x^2)$ so in the basis this vector is $(0,0,1),$ then you multiply this by the matrix and you get (-1,-1,1) which are the coefficients that you have to multiply with your basis $B$ so $-1(1,1,1)-1(1,1,0)+1(1,0,0)=(-1,-2,-1).$ To recover the transformation, use linearity. Jul 20 '20 at 0:13