I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space and let $S \subseteq \mathrm{End}_k(V)$ be a subset of pairwise commuting (i.e. $\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds:
1. If all $f \in S$ are diagonalizable, then there exist maps $\chi_i:S \rightarrow k$, $i=1,\ldots,r$, such that $V = \bigoplus_{i=1}^r E_\chi(S)$E_{\chi_i}(S)$, where$E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. 2. The maps$\chi_i$in 1 are unique. 3. 1 is equivalent to the existence of a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is diagonal. (I believe that this might not be true) 4. If all$f \in S$are trigonalizable, then there exists a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is upper triangular and for each diagonalizable$f \in S$the matrix$M_{\mathcal{B}}(f)$is diagonal. I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the$\chi$-maps when I try to prove this by induction on$\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think. Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :) One additional question: Suppose that$k$is algebraically closed and that$G$is an affine commutative algebraic group over$k$which coincides with its semisimple part, embedded as a closed subgroup in some$GL(V)$. Are the maps$\chi_i:G \rightarrow \mathbb{G}_{m}$morphisms of algebraic groups? 4 deleted 24 characters in body I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let$k$be a field, let$V$be a finite-dimensional$k$-vector space and let$S \subseteq \mathrm{End}_k(V)$be a subset of pairwise commuting (i.e.$\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds: 1. If all$f \in S$are diagonalizable, then there exist maps$\chi_i:S \rightarrow k$,$i=1,\ldots,r$, such that$V = \bigoplus_{i=1}^r E_\chi(S)$, where$E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. 2. Are the 3. The maps$\chi_i$in 1 are unique? 4. If . 5. 1 holds, then obviously there exists is equivalent to the existence of a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is diagonal. Now, is this equivalent to 1 (under the same assumptions)? I believe that this might not be true) 6. If all$f \in S$are trigonalizable, then there exists a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is upper triangular and for each diagonalizable$f \in S$the matrix$M_{\mathcal{B}}(f)$is diagonal. I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the$\chi$-maps when I try to prove this by induction on$\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think. Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :) One additional question: Suppose that$k$is algebraically closed and that$G$is an affine commutative algebraic group over$k$which coincides with its semisimple part, embedded as a closed subgroup in some$GL(V)$. Are the maps$\chi_i:G \rightarrow \mathbb{G}_{m}$morphisms of algebraic groups? 3 added 41 characters in body I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let$k$be a field, let$V$be a finite-dimensional$k$-vector space and let$S \subseteq \mathrm{End}_k(V)$be a subset of pairwise commuting (i.e.$\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds: 1. If all$f \in S$are diagonalizable, then there exist maps$\chi_i:S \rightarrow k$,$i=1,\ldots,r$, such that$V = \bigoplus_{i=1}^r E_\chi(S)$, where$E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. 2. Are the maps$\chi_i$in 1 unique? 3. If 1 holds, then obviously there exists a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is diagonal. Now, is this equivalent to 1 (under the same assumptions)? 4. If all$f \in S$are trigonalizable, then there exists a basis$\mathcal{B}$of$V$such that for each$f \in S$the matrix$M_{\mathcal{B}}(f)$of$f$with respect to$\mathcal{B}$is upper triangular and for each diagonalizable$f \in S$the matrix$M_{\mathcal{B}}(f)$is diagonal. I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the$\chi$-maps when I try to prove this by induction on$\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think. Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :) One additional question: Suppose that$k$is algebraically closed and that$G$is an affine commutative algebraic group over$k$, k$ which coincides with its semisimple part, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi_i:G \rightarrow \mathbb{G}_{m}$ morphisms of algebraic groups?