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5
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edited Dec 23 2009 at 15:06 |
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:
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 $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. The maps $\chi_i$ in 1 are unique. 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) 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?
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4
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edited Dec 22 2009 at 15:34 |
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:
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 $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. Are the The maps $\chi_i$ in 1 are unique? If . 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) 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?
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3
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edited Dec 22 2009 at 15:03 |
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:
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 $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. Are the maps $\chi_i$ in 1 unique? 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)? 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?
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2
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edited Dec 22 2009 at 14:49 |
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:
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 $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. Are the maps $\chi_i$ in 1 unique? 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)? 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 2 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). induction on $\mathrm{dim}V$). Also, I know that the first part of 34, 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 $k$, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi_i:G \rightarrow \mathbb{G}_{m}$ morphisms of algebraic groups?
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1
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asked Dec 22 2009 at 14:38 |
Simultaneous diagonalization
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:
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 $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. 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)? 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 2 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). Also, I know that the first part of 3, 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 :)
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