Let $A=b(2,\mathbb{R})$ be he Lie subalgebra of upper triangle matrices of $gl(2,\mathbb{R})$. It is clear that $e_1=(1,0)$ is an eigenvector for $A$, because it is an eigenvector for every element of $A$; that is, $a(v) \in Span\{(1,0)\}$ for every $a \in A$. Now I appreciate your answer to find the corresponding weight to $(1,0)$ and determine its weight space. Recall that we have the definition of weight as

A weight for a Lie subalgebra $A$ of $gl(V)$ is a linear map $\lambda: A \to F$ such that $V_{\lambda}= \{v \in V: av=\lambda (a) v ~~ \text{for all}~ a \in A \} $ is a non-zero subspace pf $V$.

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    $\begingroup$ Just a point to make this precise: I suppose you are looking at the defining ($2$-dimensional) representation of the Lie algebra $gl(2,\Bbb R)$ (which is then restricted to the subalgebra $A$). I might be interesting to begin writing down the definition of this representation. $\endgroup$ – Marc van Leeuwen Apr 18 at 17:24
  • $\begingroup$ I am looking at the defining (2-dimensional) representation as well, but unfortunately I do not understand the weight concept precisely! and as a consequence the concept of highest weight vectors for more studies. I appreciate your answer for a small example to clarify this concepts for me. $\endgroup$ – user40491 Apr 18 at 17:29

You seem to be hesitant to write down the definition of the action, which is simply $M.v=Mv$ (matrix times vector multiplication) for $v\in\Bbb R^2$ and $M\in gl(2,\Bbb R)$. (In your definition of $V_\lambda$, the action $M.v$ is written as $M(v)$, which I find slightly less proper as elements of the Lie algebra $gl(2,\Bbb R)$ are not functions taking a vector of some representation as argument.) Now if $M={a~~b\choose0~~c}$ and $v=e_1$ you can check that $M.v=av$, so the eigenvalue of $e_1$ for $M$ is $a$. The weight $\lambda$ of $e_1$ is the map $A\to\Bbb R$ given by $M\mapsto a$ that takes the top-left entry of the matrix; this is indeed a linear map on $A$. If you like, $\lambda$ is defined by $\lambda({a~~b\choose0~~c})=a$.

For this $\lambda$ the weight space $V_\lambda$ is $\{\,v\in V\mid\forall{a~~b\choose0~~c}\in gl(2,\Bbb R):{a~~b\choose0~~c}.v=av\,\}$ which is the span $\langle e_1\rangle$ of $e_1$ (even for a single such matrix with $b\neq0$ or $c\neq a$, the eigenspace for $a$ is just $\langle e_1\rangle$ so $V_\lambda$ is certainly contained in this subspace; and every vector $\langle e_1\rangle$ satisfies the condition for being in $V_\lambda$). It happens that this is the only weight space for this representation: most matrices in $A$ are not diagonalisable and have just this one eigenspace. (It is more customary to consider weight spaces for Abelian subalgebras of semisimple elements, such as the diagonal matrices in $gl(2,\Bbb R)$; then the weight spaces span the whole representation space.)

  • $\begingroup$ Let consider your definition of $\lambda$ then weight space is $\mathbb{R}$? $\endgroup$ – user40491 Apr 18 at 18:19
  • $\begingroup$ @user40491 No you need a subspace of $V$. See my extended answer. $\endgroup$ – Marc van Leeuwen Apr 18 at 19:44

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