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Every operator $Tin mathcal L(V)$ where $V$ is a $mathbb C-$vector space has an eigenvalue.
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Let $V$ a $mathbb C-$vector space of finite dimension (let say $n$) and $Tin mathcal L(V)$. Then $T$ has at least one eigenvalue.
I don't understand the proof.
The list $(v,T(v),...,T^n(v))$ is linearly dependent, i.e. there is $alpha _i$ not all zero s.t. $$alpha _0v+alpha _1T+...+alpha _nT^n(v)=0$$
for all $v$. Let $alpha _mneq 0$ the biggest one i.e. $alpha _k=0$ for all $k>m$. Then $$0=alpha _0v+...+alpha _mT^m(v)=c(T-lambda _1I)...(T-lambda _mI)v,$$
and thus there is $j$ s.t. $T-lambda _jI$ is not injective.
Question : I don't understand why there is $j$ s.t. $T-lambda _jI$ is not injective.
linear-algebra
asked Aug 6 at 8:48
Henri
404
add a comment |Â
up vote
0
down vote
favorite
Let $V$ a $mathbb C-$vector space of finite dimension (let say $n$) and $Tin mathcal L(V)$. Then $T$ has at least one eigenvalue.
I don't understand the proof.
The list $(v,T(v),...,T^n(v))$ is linearly dependent, i.e. there is $alpha _i$ not all zero s.t. $$alpha _0v+alpha _1T+...+alpha _nT^n(v)=0$$
for all $v$. Let $alpha _mneq 0$ the biggest one i.e. $alpha _k=0$ for all $k>m$. Then $$0=alpha _0v+...+alpha _mT^m(v)=c(T-lambda _1I)...(T-lambda _mI)v,$$
and thus there is $j$ s.t. $T-lambda _jI$ is not injective.
Question : I don't understand why there is $j$ s.t. $T-lambda _jI$ is not injective.
linear-algebra
asked Aug 6 at 8:48
Henri
404
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ a $mathbb C-$vector space of finite dimension (let say $n$) and $Tin mathcal L(V)$. Then $T$ has at least one eigenvalue.
I don't understand the proof.
The list $(v,T(v),...,T^n(v))$ is linearly dependent, i.e. there is $alpha _i$ not all zero s.t. $$alpha _0v+alpha _1T+...+alpha _nT^n(v)=0$$
for all $v$. Let $alpha _mneq 0$ the biggest one i.e. $alpha _k=0$ for all $k>m$. Then $$0=alpha _0v+...+alpha _mT^m(v)=c(T-lambda _1I)...(T-lambda _mI)v,$$
and thus there is $j$ s.t. $T-lambda _jI$ is not injective.
Question : I don't understand why there is $j$ s.t. $T-lambda _jI$ is not injective.
linear-algebra
asked Aug 6 at 8:48
Henri
404
Let $V$ a $mathbb C-$vector space of finite dimension (let say $n$) and $Tin mathcal L(V)$. Then $T$ has at least one eigenvalue.
I don't understand the proof.
The list $(v,T(v),...,T^n(v))$ is linearly dependent, i.e. there is $alpha _i$ not all zero s.t. $$alpha _0v+alpha _1T+...+alpha _nT^n(v)=0$$
for all $v$. Let $alpha _mneq 0$ the biggest one i.e. $alpha _k=0$ for all $k>m$. Then $$0=alpha _0v+...+alpha _mT^m(v)=c(T-lambda _1I)...(T-lambda _mI)v,$$
and thus there is $j$ s.t. $T-lambda _jI$ is not injective.
Question : I don't understand why there is $j$ s.t. $T-lambda _jI$ is not injective.
linear-algebra
asked Aug 6 at 8:48
Henri
404
asked Aug 6 at 8:48
Henri
404
asked Aug 6 at 8:48
Henri
404
asked Aug 6 at 8:48
Henri
404
404
add a comment |Â
add a comment |Â
1 Answer
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Product of a finite number of injective linear maps is injective. Since $0$ is not injective it follows that one of the factors on the right is not injective.
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Product of a finite number of injective linear maps is injective. Since $0$ is not injective it follows that one of the factors on the right is not injective.
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
2
down vote
accepted
Product of a finite number of injective linear maps is injective. Since $0$ is not injective it follows that one of the factors on the right is not injective.
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Product of a finite number of injective linear maps is injective. Since $0$ is not injective it follows that one of the factors on the right is not injective.
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
Product of a finite number of injective linear maps is injective. Since $0$ is not injective it follows that one of the factors on the right is not injective.
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
answered Aug 6 at 8:51
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
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