# Matrix of a Linear Operator (Change of Basis)

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*}

• What is the basis? $\mathcal{B} = \{x^2-x,1,x^2+2\}$ or $\mathcal{B} = \{x^2-x,1,x^2+x\}$? Dec 30, 2023 at 23:54
• Sorry it should have been $x^2+x$. I corrected it in the edit. Dec 30, 2023 at 23:59
• 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. Dec 31, 2023 at 0:15
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 $$[D]_\mathcal{B}=\begin{bmatrix}-1 & 0 & -1 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix}.$$