Mixing
Proof about an endomorphism of a finite-dimensional vector space
Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,
$$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$
I don't know exactly how to start this proof. Could you help me? Thanks in advance!
linear-algebra
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
add a comment |
Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,
$$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$
I don't know exactly how to start this proof. Could you help me? Thanks in advance!
linear-algebra
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
add a comment |
Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,
$$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$
I don't know exactly how to start this proof. Could you help me? Thanks in advance!
linear-algebra
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
Let $f in operatorname{End}(E)$. Prove that, if $E$ has finite dimension,
$$exists m in mathbb{N} quad s.t.quad operatorname{Im}f^n=operatorname{Im}f^m quad and quad ker f^n = ker f^m quad forall n ge m$$
I don't know exactly how to start this proof. Could you help me? Thanks in advance!
linear-algebra
linear-algebra
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 18:07
Monstrous Moonshiner
2,25011337
2,25011337
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
asked Nov 21 '18 at 17:40
GibbsGibbs
13110
13110
add a comment |
add a comment |
1 Answer
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First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.
The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
add a comment |
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First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.
The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
add a comment |
First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.
The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
add a comment |
First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.
The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
First, you have to see that $operatorname{Im} f^n subseteq operatorname{Im} f^m$ and $ker f^m subseteq ker f^n$ whenever $m le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $operatorname{Im} f^m$, and being subspaces of $operatorname{Im} f^m$, they must also be equal to $operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.
The second part proceeds almost exactly the same way. Since ${ker f^m}_{m=0}^infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $ker f^n$ has the same dimension as $ker f^m$ for all $n ge m$, and since $ker f^n$ contains $ker f^m$ and has the same dimension, we must have $ker f^n = ker f^m$, as expected.
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
edited Nov 21 '18 at 18:54
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
answered Nov 21 '18 at 18:03
Monstrous MoonshinerMonstrous Moonshiner
2,25011337
2,25011337
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
add a comment |
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right?
– Gibbs
Nov 21 '18 at 18:43
1
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 '18 at 18:54
add a comment |
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