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Are "root vectors" and generilised eigenvectors the same thing? Im reading a book written by Nikolski named "Operators, Functions, and Systems - An Easy Reading: Hardy, Hankel, and Toeplitz" and he uses the term every now and then.

A generilised eigenvector is a vector $v$ such that

$(A-\lambda)^{n}v=0$ for some $n$ and $(A-\lambda)^{k}\ne 0$ for $k

Looking for references.

2 Answers 2

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A root vector in this context is a generlized eigenvector.

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    You give yourselves the green chevron although your post was downvoted. Are you a joker? – 2018-02-14
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    @loupblanc right is right. People can vote however they want. I added the name of the book. – 2018-02-14
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    You ask the question: "root vector=generalized eigenvector" ? and you answer by yes. I think that you are right; yet, it is not for you to judge yourselves. In fact, I think that the OP should not have the right to judge himself. – 2018-02-14
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    @loup blanc well I asked an old professor who knew about this termonolgy. He did not have a stack accont :) – 2018-02-15
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No. Given a Lie algebra $L$ and a Cartan subalgebra $H$, a root is a non-zero element $\alpha\in H^*$ such that the associated root space $L_{\alpha}=\left\{x\in L\mid [h,x]=\alpha(h)x, \forall h\in H\right\}$ is non-zero. A non-zero $x\in L_{\alpha}$ is called a root vector.

You could see a root $\alpha$ as generalisation of an eigenvalue in the sense that it satisfies the equation $[h,x]=\alpha(h)x$. Here $x$ is an eigenvector of $[h,-]$ with eigenvalue $\alpha(h)$, but unlike standard linear algebra, this holds for all $h\in H$, hence $x$ is simultaneously an eigenvector for all $[h,-]$.

The importance of roots is that given a semisimple complex Lie algebra $L$, you can decompose $L$ as $$L=L_0\oplus \bigoplus_{\alpha\in \Phi}L_{\alpha},$$ where $\Phi$ is the set of roots. It turns out that $L_0=H$ is a Cartan subalgebra of $L$. The set $\Phi$ is called the root system of $L$. One can associate a Dynkin diagram to the root system, and these diagrams are classified. It follows that the root systems can be classified which in turn can be used to classify the semisimple Lie algebras. This is perhaps one of greatest achievements in mathematics and is not very difficult to understand. (In fact, you only need basic linear algebra to understand this).

Generalized eigenvectors on the other hand are very different from elements in root spaces. Given a linear transformation $T:V\rightarrow V$ on a finite-dimensional vector space (over an algebraically closed field such as $\mathbb{C})$, one can choose a basis of $V$ consisting of eigenvectors and generalized eigenvectors. Considering the matrix of $T$ w.r.t. this basis yields he Jordan canonical form of $T$. When $T$ is diagonalizable, there are only eigenvectors in this basis and no (strict) generalized eigenvectors.

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    I found this ; https://www.encyclopediaofmath.org/index.php/Root_vector ,but only references in french. I think root system are different things. sorry about not posting this but I try to avoid links questions. – 2017-01-21
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    Hmm, I haven't read the book by Nikolski. If the book doesn't mention Lie algebras, root vectors are probably not going to be what I explained. It appears that the link you gave uses the word root vector for generalized eigenvectors. I have never encountered this practice, but as long as you don't use the words in Lie algebra theory you should be fine. – 2017-01-21
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    His book is on analysis and not algebra. He didnt define it tho, I wouder if im missing something but it loooks like it is the same thing. – 2017-01-21
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    Yeah, he probably means generalized eigenvector. I guess you're dealing with operators then. I don't think it is standard practice to use the word root vector in this context, but you can do it. Well at least you got a brief overview of Lie algebra theory :D – 2017-01-21
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    Bit late to the party, but should anyone else stumble on this and want to understand root vectors and roots in the context of root systems these notes http://www-f1.ijs.si/~ziherl/Susic11.pdf are great. (particularly if you are coming t this from a physics PoV) – 2017-05-06
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    One should add what is a root *vector* in this context: it is an element of the root space, the one you denote by $x$. – 2018-02-14
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    @მამუკა ჯიბლაძე: Correct I never explained what a root *vector* is. I added it to the explanation (which unfortunately turns out to be useless with regard to this question). – 2018-02-14
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    Well I believe it is useful at least in that it warns about unfortunate terminology clash. – 2018-02-14