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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

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$[T]_{B,A}$ 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 $T(1+x^2)$?

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  • $\begingroup$ how did you get T(2-3x)=(0,0,1)? I multiplied T with (2,-3,0) to get (-1,3,2) which becomes (0,2,-1)? $\endgroup$
    – John Rawls
    Jul 19, 2020 at 23:53
  • $\begingroup$ I multiplied $[T]$ by $(0,1,0).$ You are multiplying by the basis $\{1,x,x^2\}$ but the basis here is $\{1,2-3x,1+x^2\}$ $\endgroup$
    – Phicar
    Jul 19, 2020 at 23:55
  • $\begingroup$ im still confused, so then to find T(1+x^2) we'd need to multiply by (0,0,1)? where are these vectors coming from? $\endgroup$
    – John Rawls
    Jul 19, 2020 at 23:58
  • $\begingroup$ and after we get all of the resulting vectors as you suggest how does that get us the formula of the linear transformation? $\endgroup$
    – John Rawls
    Jul 20, 2020 at 0:00
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    $\begingroup$ 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. $\endgroup$
    – Phicar
    Jul 20, 2020 at 0:13

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