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If $$ A = \begin{pmatrix} 1 & -1 & 2 \\ -2 & 1 &-1 \\ 1 & 2 & 3 \end{pmatrix} $$ is the matrix representation of a linear transformation $T : P_3(x) \to P_3(x)$ with respect to bases $\{1-x,x(1-x),x(1+x)\}$ and $\{1,1+x,1+x^2\}$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.

$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$

$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$

$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$

Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$

Is this fine ?

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Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

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What you did is fine, but now you have to compute $T[\alpha+\beta x+\gamma x^2]$ for arbitrary $\alpha,\beta,\gamma\in\mathbb R$. In order to do that, solve the equation$$\alpha+\beta x+\gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$\left\{\begin{array}{l}a-b+2c=\alpha\\-2a+b-c=\beta\\a+2b+3c=\gamma.\end{array}\right.$$

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  • $\begingroup$ then it becomes a $ T:[{(1,x,x^2)} -> {(1,1+x,1+x^2)}]$ ? $\endgroup$ – Sarkar Jul 21 '18 at 16:12
  • $\begingroup$ This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$. $\endgroup$ – José Carlos Santos Jul 21 '18 at 16:19
  • $\begingroup$ Okay. i get it. So now the coefficients of basis ${(1,1+x,1+x^2)}$ is standardised using standard basis${(1,x,x^2)}$ coefficients $\endgroup$ – Sarkar Jul 21 '18 at 16:29
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I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),\mathcal B_1)$ to $(P_2(x),\mathcal B_2)$, i.e. we have a matrix relation $$Y_2=AX_1$$ and asked for the matrix of this same linear transformation from $(P_2(x),\mathcal B_\text{canon})$ to itself, i.e. we're asked for the matrix $T$ such that $$Y=TX.$$ Now that's easy, given the change of basis matrices: $$P_1=\begin{bmatrix}\!\!\begin{array}{rrc} 1&0&0\\-1&1&1\\0&\:\llap-1&1 \end{array}\end{bmatrix},\qquad P_2=\begin{bmatrix} 1&1&1\\0&1&0\\0&0&1 \end{bmatrix}.$$ We have $Y=P_2Y_2$, $X=P_1X_1$, so $$Y=P_2Y_2=P_2AX_1=(\underbrace{P_2AP_1^{-1}}_T)X.$$ There remains to find the inverse of $P_1$, which is standard by row reduction.

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