Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

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


From here you should be home free.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.