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I need to find the trace of differentiation operator $D$ on a polynomial vector space $P$ with degree $n$. $Dp(x)=p'(x)$. According to wikipedia trace can be found by representing the basis in matrix form. Basis for this will be $\{1,x,x^2,\dots,x^n\}$. Now how can $D$ be defined as a matrix relative to this basis so that trace can be found?

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How do you usually find a matrix of a linear transformation relative to a basis? I would start by considering what $Dx^k$ is in terms of the basis. –  Jonas Meyer Feb 19 '12 at 3:38
    
Can you point to some example? I am not sure how to do it –  codejammer Feb 19 '12 at 3:57
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Let $\vec e_k=x^k$ for $k=0,1,\dots,n$; these $\vec e_k$ are your basis vectors. A polynomial $$p(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$$ can then be written $$p=a_0\vec e_0+a_1\vec e_1+\dots+a_n\vec e_n$$ as a linear combination of the basic vectors. In other words, in terms of this basis you can think of $p$ as being represented by $$\left[\matrix{a_0\\a_1\\\vdots\\a_n}\right]\in\mathbb{R}^{n+1}\;.$$ Now what does the transformation $D$ do to $p$?

$$\begin{align*} Dp(x)&=p'(x)\\ &=a_1+a_2(2x)+a_3(3x^2)+\dots+a_n(nx^{n-1})\\ &=a_1D(\vec e_1)+a_2D(\vec e_2)+a_3D(\vec e_3)+\dots+a_nD(\vec e_n)\\ &=a_1\vec e_0+a_2(2\vec e_1)+a_3(3\vec e_2)+\dots+a_n(n\vec e_{n-1})\;. \end{align*}$$

If we replace the $\vec e_k$ by their representations in $\mathbb{R}^{n+1}$ with respect to our basis, this becomes

$$a_0\left[\matrix{0\\0\\0\\\vdots\\0\\0}\right]+a_1\left[\matrix{1\\0\\0\\\vdots\\0\\0}\right]+a_2\left[\matrix{0\\2\\0\\\vdots\\0\\0}\right]+a_3\left[\matrix{0\\0\\3\\\vdots\\0\\0}\right]+\dots+a_n\left[\matrix{0\\0\\0\\\vdots\\n\\0}\right]\;,$$ which can be rewritten as

$$\left[\matrix{0&1&0&0&\dots&0\\0&0&2&0&\dots&0\\0&0&0&3&\dots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\dots&n\\0&0&0&0&\dots&0}\right]\left[\matrix{a_0\\a_1\\a_2\\a_3\\\vdots\\a_n}\right]\;.$$

From here you should be home free.

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