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What is the null space of differentiation transformation: $\frac{\mathrm{d} }{\mathrm{d} x}:P_{n} \to P_{n}$ where $P_{n}$ is the space of all polynomials of degree $\leqslant n $ over the real numbers What is the null space of the second derivative as a transformation of $P_{n}$ ? What is the null space of the kth derivative?

I am slightly at a loss here, as I realise that they are looking differentiation as a transformation, but not a simple algebraic one as that(e.g reflection over x axis or something like that, which can easily formulated into a matrix format). Can anyone :

  1. Give me some hints on how to approach this problem, specifically, express the differentiation as a matrix?
  2. Point me to some text/textbook which can help me build such concepts in a better way.

Further Edit: $P_{n}$ is as rightly pointed out is a vector space made of linear combination of the basis set {$1,x,x^{2},...,x^{n}$}. So I reckon the y=d/dx p(x) = AX where,

$p'(x)=0$ for the null space of the transformation operator.

So one trivial solution is when p(x)=c

Till now, this is what I could figure.

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Start with $n=3$.... –  user38268 May 28 '12 at 9:51
should I just treat d/dx as an operator and not as a matrix? Because I knew nullspace only in the context of homogenous linear systems –  Soham May 28 '12 at 9:54
You can treat it as a matrix if you like, but in this case it doesn't really help you find the null space, at least not computationally: it might help you better understand linear maps, though. –  Ben Millwood May 28 '12 at 10:04
Please don't "sign" your posts with your name (not even once, certainly not multiples times). See the FAQ. Your name already appears on the bottom right of all your posts. –  Arturo Magidin May 28 '12 at 20:42
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4 Answers

up vote 5 down vote accepted

As a linear operator on $P_n$, differentiation can be expressed in matrix terms, but doing so is completely unnecessary and obscures what’s really going on; you’re better off thinking in terms of the definition of the null space of a linear transformation.

The null space of an operator $T:V\to W$ is simply the set $\{v\in V:T(v)=0_W\}$ of vectors in $V$ that get sent to the $0$ vector of $W$ by $T$. For what polynomials $p(x)\in P_n$ is it true that $$\frac{d}{dx}p(x)$$ is the zero vector of $P_n$? (For starters, what is the zero vector of $P_n$?) This has a very simple answer that requires no fiddling with matrices.

Once you’ve handled the first derivative, the rest should be easy. To find the null space of $\frac{d^3}{dx^3}$, for instance, just ask yourself which $p(x)\in P_n$ have the property that $$\frac{d^3}{dx^3}p(x)$$ is the zero vector of $P_n$.

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yes I think I get it. Let $p(x) = a_{1}x^{n} + a_{2}x^{n-1} +...+a_{n} $ => p'(x)=0 One trivial solution is when p(x) = c . Do let me know, if I am completely off track. –  Soham May 28 '12 at 10:22
I think you're going in the right direction. –  Eric Stucky May 28 '12 at 10:24
@Soham: Looks good to me! –  Brian M. Scott May 28 '12 at 10:25
@Brain M Scott Thanks, I think I can take it from here! Thanks a lot, Sir! It was very very helpful. Can you still point me to some foundation text.I intend to really play around this concept in my mind. I am using Erwin Kreyszig's book. Your help will be much appreciated. Further can you point me to some text where they have transformed d/dx as a matrix. I have scribbled an equivalent matrix on paper, would like to check if its right –  Soham May 28 '12 at 10:28
@Soham: When I retired, I gave away most of my linear algebra texts, so I can’t really offer good advice there. For the matrix itself, if $p(x)=a_nx^n+\ldots+a_0$ is represented by the vector of its coefficients, you get this: $$\pmatrix{0&0&0&\dots&0&0\\n&0&0&\dots&0&0\\0&n-1&0&\dots&0&0\\0&0&n-2&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\dots&1&0}\pmatrix{a_n\\a_{n-1}\\a‌​_{n-2}\\a_{n-3}\\\vdots\\a_1\\a_0}=\pmatrix{0\\na_n \\ (n-1)a_{n-1} \\ (n-2)a_{n-2}\\ \vdots\\2a_2\\a_1}$$ –  Brian M. Scott May 28 '12 at 10:38
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For general function spaces, the differentiation operator may be difficult to put into a matrix, but consider that $a_n x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ becomes $(n)a_nx^{n-1}+(n-1)a_nx^{n-2}+\dots+(1)a_1+0$.

If you consider the initial polynomial to be the vector $[a_n, a_{n-1},\dots, a_1, a_0]^T$, there is actually a nice matrix representation for what just happened to it. Like Lim, I would encourage you to use small examples first.

Alternatively, don't forget the definition of nullspace of $D$ is all the elements $p(x)$ such that $D(p(x))=0$.

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Thanks a lot man, your last statement was very helpful. Its just this, I knew this idea but only on the surface, your comment did help me a lot. –  Soham May 28 '12 at 10:30
Thanks; it's always nice to hear. –  Eric Stucky May 28 '12 at 10:30
I am not done yet, suggest me a book, and I would like to check my answer of d/dx (as a matrix), so please point me to some text as well which represents so. –  Soham May 28 '12 at 10:33
I personally learned linear algebra from Gilbert Strang's $Introduction$ $to$ $Linear$ $Algebra$, but there are probably much cheaper alternatives. –  Eric Stucky May 28 '12 at 10:42
Aah, thats a classic. –  Soham May 28 '12 at 14:07
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To write differentiation in $P_n$ as a matrix, first establish a basis for $P_n$. There are several (infinitely many, of course) but the most usual one is the basis $\left\{1,x,x^2,\dots,x^n\right\}$. Then the matrix of differentiation is an $(n+1)\times (n+1)$ matrix (notice that there are $n+1$ elements in the basis, so $P_n$ is actually $(n+1)$-dimensional), with entries $A_{ij}$ such that differentiating the $j^\mathrm{th}$ basis vector (i.e. $x^{j-1}$) produces $A_{ij}$ of the $i^\mathrm{th}$ basis vector (so for example since differentiating $x^2$ doesn't produce any $x^4$, we have $A_{42}=0$).

Edit: some of the other answerers definitely have a point when they say working out the null space of differentiation doesn't require matrices at all (thinking about linear maps in the abstract is sufficient and often simpler), but nevertheless if you want to see how to make one, this is what you have to do.

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Thanks a lot, I think, one of the first concept it had to strike was view $P_{n}$ as a vector space {which I verified in my mind as being associative under addition operator, and multiplication by scalar operator}, but I failed to see it as having a basis of ${1,x,x^{2},...,x^{n}}$. But the biggest concept builder was viewing it has an (n+1) x (n+1) matrix. –  Soham May 28 '12 at 10:11
I sort of skipped over those details because I assumed you'd know them already; it might have been better to ask the question "how do I view $P_n$ as a vector space". But then, sometimes finding the right question is at least as hard as finding the answer to it :) –  Ben Millwood May 28 '12 at 10:15
Thanks a lot man, your idea about the differentiation matrix has got me perked up, can you recommend me a book or something. Once again, thanks a lot! –  Soham May 28 '12 at 10:31
No problem, but I don't have any good linear algebra books – my lecture notes were very thorough but aren't online :( –  Ben Millwood May 28 '12 at 21:01
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A polynomial $a_0+a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}$ living in $P_n$ can also be viewed as a vector $(a_0,a_1,\dots,a_{n-1})$ living in ${\bf R}^n$.

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Thanks a lot man, appreciate your time,effort and care for helping me out. Thanks –  Soham May 28 '12 at 10:30
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