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Let $X$ be the vector space of polynomials $p(x)$ having real coefficients and degree at most 2. Consider the basis $\mathcal{B} = \{x^2-x,1,x^2+x\}$ of $X$. The differentiation operator $D: X \to X$ given by $Dp(x) = p'(x)$ is linear. What is the matrix of $D$ with respect to $\mathcal{B}$?

It's been a while, so I'm just writing this to ask for a check. First, I will calculate the image of each basis vector under the transformation. We have,

\begin{align*} D(x^2-x) &= 2x-1\\ D(1) &= 0\\ D(x^2+x) &= 2x+1. \end{align*}

To find $\lbrack D \rbrack_{\mathcal{B}}$, I need to find the coordinate vector of the image of each of these basis vectors. We can find these by solving the linear systems, \begin{align*} \begin{bmatrix} 1 & 0 & 1 & 0\\ -1 & 0 & 1 & 2\\ 0 & 1 & 0 & -1\\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 &-1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}\\ &\vdots \end{align*} This results in the vectors $(-1 \ \ -1 \ \ 1)$, $(0 \ \ 0 \ \ 0)$, and $(-1 \ \ 1 \ \ 1)$, so the matrix is,

\begin{align*} \lbrack D \rbrack_{\mathcal{B}} =\begin{bmatrix} -1 & 0 & -1\\ -1 & 0 & 1\\ 1 & 0 & 1\\ \end{bmatrix}. \end{align*}

Am I thinking about this correctly?

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  • $\begingroup$ What is the basis? $\mathcal{B} = \{x^2-x,1,x^2+2\}$ or $\mathcal{B} = \{x^2-x,1,x^2+x\}$? $\endgroup$ Dec 30, 2023 at 23:54
  • $\begingroup$ Sorry it should have been $x^2+x$. I corrected it in the edit. $\endgroup$ Dec 30, 2023 at 23:59
  • $\begingroup$ Write $u=x^2+x, v=1, w=x^2-x$. This means $D(u)=u-w-v$ which should be shown in the matrix. Use this idea to check the other results as well. $\endgroup$ Dec 31, 2023 at 0:15

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Your result is correct. Note that \begin{align} &D(x^2-x)=(-1)(x^2-x)+(-1)(1)+1(x^2+x);\\ &D(1)=0(x^2-x)+0(1)+0(x^2+x);\\ &D(x^2+x)=(-1)(x^2-x)+1(1)+1(x^2+x). \end{align} Therefore, the matrix of $D$ with respect to $\mathcal{B}=\{x^2-x,1,x^2+x\}$ is given by $$[D]_\mathcal{B}=\begin{bmatrix}-1 & 0 & -1 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix}.$$

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    $\begingroup$ Thanks for the check. Your way is much more quick and elegant! $\endgroup$ Dec 31, 2023 at 0:21

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