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What does the following mean --

"The Jordan Canonical Form of the operator $w{d\over dw}$ acting on the complex vector space of polynomials in $w$ of degree less than $n$"?

Thank you.

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Which part do you not understand? – Chris Eagle May 19 '12 at 17:48
@ChrisEagle: I understand what a complex vector space of polynomials in $w$ of degree less than $n$ means, but not the bit before it. I know how a JCF matrix looks like, but how do you turn an operator into a matrix?? – HgAu May 19 '12 at 17:50
$w\frac{d}{dw}$ is a linear map on a comple vector space, the composition of multiplication by $w$ and $\frac{d}{dw}$ so it has a Jordan form, it is a sum of a semisimple linear map and a nilpotent, and in an appropriate basis this is called a Jordan normal form. I guess you should rather ask what that form is. It is probably something very close to the identity, when you differentiate you lower by one power of $w$ and multiply by an integer, so you revert part of that multiplying by $w$. – plm May 19 '12 at 18:04
I should have said "Jordan decomposition" instead of "Jordan form", or "Jordan-Chevalley decomposition". This refers to the sum decomposition. The "normal form" is actually writing the corresponding triangular matrix in a basis of generalized eigenvectors. – plm May 19 '12 at 18:18

There is a natural basis for our space, namely the $w^k$. And it is easy to write down the matrix of the operator with respect to that basis. What does $w\frac{d}{dw}$ do to $w^k$?

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Thank you for answering! $w^k$ becomes $kw^k$, but I am still not sure what the matrix looks like. So each column is the image of the basis vector it represents? So $(0\,\,\,w\,\,\,2w^2...(n-1)w^{n-1})$? – HgAu May 19 '12 at 18:00
Wait... that isn't in canonical form... how can I make it into canonical form? – HgAu May 19 '12 at 18:02
You have the idea (columns) right. The first column is all $0$'s. The second column is $0,1,0,\dots$. The third is $0,0,2,0,\dots$. And so on. OK, we have the matrix. Now if it is not in canonical form, put it in canonical form. – André Nicolas May 19 '12 at 18:08
Ah, thanks! I was confused because I thought they are all squashed into one row -- silly me – HgAu May 19 '12 at 18:16

A very obvious basis of polynomials of degree n in $w$ is $1,w,...,w^n$. They map to $0,w,2w^2,3w^3,...,nw^n$ under $P(w)\rightarrow w\frac{dP(w)}{dw}$. So this basis is actually made of eigenvector, that is, your linear map is diagonal with $0,1,2,...,n$ diagonal, and in particular already in Jordan normal/canonical form.

For $n=2$. \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}

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Thank you for answering, plm! – HgAu May 19 '12 at 18:22

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