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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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    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.2012-05-19

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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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    Ah, thanks! I was confused because I thought they are all squashed into one row -- silly me2012-05-19
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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}