Functorial proof of Cayley-Hamilton using exterior powers
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$begingroup$
Let $V$ be a rank $n$ free module over a commutative ring $R$. Let $dagger$ denote the adjoint with respect to the natural perfect pairing given by the wedge product $$Lambda ^kotimes Lambda ^{n-k}overset{wedge}{longrightarrow} Lambda ^n.$$
Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove $(Lambda ^{n-k}f)^daggercirc Lambda ^kf=det fcdot 1_{Lambda^kV}$.
Now let $Voverset{f}{to}V$ be an $R$-linear endomorphism. It induces an $R[f]$-module structure on $V$ which in turn induces an $R[f,t]$-module structure on the $R[t]$-module $Votimes _RR[t]$. Define the characteristic polynomial $chi_fin R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $Votimes _RR[t]$.
By the above fact we have the following equation in the category of $R[t]$-modules. $$(Lambda ^{n-k}(f-t))^daggercirc Lambda ^k(f-t)=chi_fcdot 1_{Lambda^k(Votimes_RR[t])}$$
I'm trying to follow the proof of Cayley-Hamilton along these lines given in 28.10, but I am confused by the sudden passage to the category of $R[f,t]cong R[f]otimes _RR[t]$-modules.
How to formally derive the Cayley-Hamilton theorem from the latter equation?
linear-algebra commutative-algebra modules exterior-algebra cayley-hamilton
asked Jan 16 at 23:23
ArrowArrow
5,19431446
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add a comment |
$begingroup$
Let $V$ be a rank $n$ free module over a commutative ring $R$. Let $dagger$ denote the adjoint with respect to the natural perfect pairing given by the wedge product $$Lambda ^kotimes Lambda ^{n-k}overset{wedge}{longrightarrow} Lambda ^n.$$
Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove $(Lambda ^{n-k}f)^daggercirc Lambda ^kf=det fcdot 1_{Lambda^kV}$.
Now let $Voverset{f}{to}V$ be an $R$-linear endomorphism. It induces an $R[f]$-module structure on $V$ which in turn induces an $R[f,t]$-module structure on the $R[t]$-module $Votimes _RR[t]$. Define the characteristic polynomial $chi_fin R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $Votimes _RR[t]$.
By the above fact we have the following equation in the category of $R[t]$-modules. $$(Lambda ^{n-k}(f-t))^daggercirc Lambda ^k(f-t)=chi_fcdot 1_{Lambda^k(Votimes_RR[t])}$$
I'm trying to follow the proof of Cayley-Hamilton along these lines given in 28.10, but I am confused by the sudden passage to the category of $R[f,t]cong R[f]otimes _RR[t]$-modules.
How to formally derive the Cayley-Hamilton theorem from the latter equation?
linear-algebra commutative-algebra modules exterior-algebra cayley-hamilton
asked Jan 16 at 23:23
ArrowArrow
5,19431446
$endgroup$
add a comment |
$begingroup$
Let $V$ be a rank $n$ free module over a commutative ring $R$. Let $dagger$ denote the adjoint with respect to the natural perfect pairing given by the wedge product $$Lambda ^kotimes Lambda ^{n-k}overset{wedge}{longrightarrow} Lambda ^n.$$
Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove $(Lambda ^{n-k}f)^daggercirc Lambda ^kf=det fcdot 1_{Lambda^kV}$.
Now let $Voverset{f}{to}V$ be an $R$-linear endomorphism. It induces an $R[f]$-module structure on $V$ which in turn induces an $R[f,t]$-module structure on the $R[t]$-module $Votimes _RR[t]$. Define the characteristic polynomial $chi_fin R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $Votimes _RR[t]$.
By the above fact we have the following equation in the category of $R[t]$-modules. $$(Lambda ^{n-k}(f-t))^daggercirc Lambda ^k(f-t)=chi_fcdot 1_{Lambda^k(Votimes_RR[t])}$$
I'm trying to follow the proof of Cayley-Hamilton along these lines given in 28.10, but I am confused by the sudden passage to the category of $R[f,t]cong R[f]otimes _RR[t]$-modules.
How to formally derive the Cayley-Hamilton theorem from the latter equation?
linear-algebra commutative-algebra modules exterior-algebra cayley-hamilton
asked Jan 16 at 23:23
ArrowArrow
5,19431446
$endgroup$
Let $V$ be a rank $n$ free module over a commutative ring $R$. Let $dagger$ denote the adjoint with respect to the natural perfect pairing given by the wedge product $$Lambda ^kotimes Lambda ^{n-k}overset{wedge}{longrightarrow} Lambda ^n.$$
Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove $(Lambda ^{n-k}f)^daggercirc Lambda ^kf=det fcdot 1_{Lambda^kV}$.
Now let $Voverset{f}{to}V$ be an $R$-linear endomorphism. It induces an $R[f]$-module structure on $V$ which in turn induces an $R[f,t]$-module structure on the $R[t]$-module $Votimes _RR[t]$. Define the characteristic polynomial $chi_fin R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $Votimes _RR[t]$.
By the above fact we have the following equation in the category of $R[t]$-modules. $$(Lambda ^{n-k}(f-t))^daggercirc Lambda ^k(f-t)=chi_fcdot 1_{Lambda^k(Votimes_RR[t])}$$
I'm trying to follow the proof of Cayley-Hamilton along these lines given in 28.10, but I am confused by the sudden passage to the category of $R[f,t]cong R[f]otimes _RR[t]$-modules.
How to formally derive the Cayley-Hamilton theorem from the latter equation?
linear-algebra commutative-algebra modules exterior-algebra cayley-hamilton
linear-algebra commutative-algebra modules exterior-algebra cayley-hamilton
asked Jan 16 at 23:23
ArrowArrow
5,19431446
asked Jan 16 at 23:23
ArrowArrow
5,19431446
asked Jan 16 at 23:23
ArrowArrow
5,19431446
asked Jan 16 at 23:23
ArrowArrow
5,19431446
asked Jan 16 at 23:23
ArrowArrow
5,19431446
5,19431446
add a comment |
add a comment |
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