2
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.