What is a root vector? 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 $k<n$.
Looking for references.
 A: 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.
A: A root vector in this context is a generlized eigenvector.
