Proof about an endomorphism of a finite-dimensional vector space
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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 at 18:07
Monstrous Moonshiner
2,21911337
asked Nov 21 at 17:40
Gibbs
485
add a comment |
up vote
2
down vote
favorite
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 at 18:07
Monstrous Moonshiner
2,21911337
asked Nov 21 at 17:40
Gibbs
485
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 at 18:07
Monstrous Moonshiner
2,21911337
asked Nov 21 at 17:40
Gibbs
485
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 at 18:07
Monstrous Moonshiner
2,21911337
asked Nov 21 at 17:40
Gibbs
485
edited Nov 21 at 18:07
Monstrous Moonshiner
2,21911337
asked Nov 21 at 17:40
Gibbs
485
edited Nov 21 at 18:07
Monstrous Moonshiner
2,21911337
edited Nov 21 at 18:07
Monstrous Moonshiner
2,21911337
edited Nov 21 at 18:07
Monstrous Moonshiner
2,21911337
2,21911337
asked Nov 21 at 17:40
Gibbs
485
asked Nov 21 at 17:40
Gibbs
485
asked Nov 21 at 17:40
Gibbs
485
485
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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 at 18:03
Monstrous Moonshiner
2,21911337
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 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 at 18:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 at 18:03
Monstrous Moonshiner
2,21911337
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 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 at 18:54
add a comment |
up vote
1
down vote
accepted
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 at 18:03
Monstrous Moonshiner
2,21911337
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 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 at 18:54
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 at 18:03
Monstrous Moonshiner
2,21911337
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 at 18:03
Monstrous Moonshiner
2,21911337
edited Nov 21 at 18:54
answered Nov 21 at 18:03
Monstrous Moonshiner
2,21911337
answered Nov 21 at 18:03
Monstrous Moonshiner
2,21911337
answered Nov 21 at 18:03
Monstrous Moonshiner
2,21911337
2,21911337
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 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 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 at 18:43
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 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 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 at 18:43
1
1
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 at 18:54
@Gibbs That is correct
– Monstrous Moonshiner
Nov 21 at 18:54
add a comment |
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