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Here is my problem:

Let $V$ be the vector space of polynomials of degree up to $2$.

and $T:V\rightarrow V$ be a linear transformation defined by the type:

$T(p(x))=p(2x+1)$

Find the matrix form of this linear transformation. The base to find the matrix is $B=\{1,x,x^2\}$

I have seen in the past other exercises about finding the matrix of a linear transformation but they were all simpler and I have no idea how to proceed with this one. Any help will be appreciated.

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  • $\begingroup$ To find a matrix, you need a basis $\endgroup$ – J. W. Tanner Sep 11 at 10:02
  • $\begingroup$ @J.W.Tanner Yes you are right. The exercise said to use the "usual basis" $B=\{1,x,x^2\}$ $\endgroup$ – michail vazaios Sep 11 at 10:05
  • $\begingroup$ How did you solve other similar exercises? Did you try doing precisely the same thing for this problem? Where did you get stuck? $\endgroup$ – HerrWarum Sep 11 at 10:06
  • $\begingroup$ @HerrWarum I've solved similar exercises where the type of the linear transformation was given explicitly. It was given like $T((x,y)^T)=(2x+y,x-y)^T$. Where I got stuck is that I can't the type(or such a type for this transformation). $\endgroup$ – michail vazaios Sep 11 at 10:12
  • $\begingroup$ To work out what this transformation does to the basis vectors, simply substitute $2x+1$ for $x$ in them. That’s what the transformation does. $\endgroup$ – amd Sep 11 at 17:17
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This means $T(ax^2+bx+c)=(4ax^2+(4a+2b)x+a+b+c)$. Taking the bais as $(x^2,x,1)$, we can write $$T \begin{pmatrix} a \\ b \\c \end{pmatrix}= \begin{pmatrix} 4a \\ 4a+2b \\a+b+c \end{pmatrix} \Rightarrow \begin{pmatrix} 4 & 0 & 0 \\ 4 & 2 & 0 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} a \\ b \\c \end{pmatrix}.$$ So the required $T$ matrix is $$T= \begin{pmatrix} 4 & 0 & 0 \\ 4 & 2 & 0 \\ 1 & 1 & 1 \end{pmatrix}.$$

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