Equivalence of surjectivity and injectivity for linear operators on finite dimensional vector spaces
$begingroup$
I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by
Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$
Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.
$w = v^i(T e_i) = v^ie'_i$.
Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.
Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.
Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?
linear-transformations
asked Jul 16 '16 at 11:36
user2582713user2582713
136
$endgroup$
|
show 7 more comments
$begingroup$
I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by
Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$
Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.
$w = v^i(T e_i) = v^ie'_i$.
Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.
Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.
Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?
linear-transformations
asked Jul 16 '16 at 11:36
user2582713user2582713
136
$endgroup$
$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26
$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27
$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12
$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15
$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18
|
show 7 more comments
$begingroup$
I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by
Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$
Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.
$w = v^i(T e_i) = v^ie'_i$.
Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.
Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.
Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?
linear-transformations
asked Jul 16 '16 at 11:36
user2582713user2582713
136
$endgroup$
I'd like to show that for a linear operator $T$ and finite-dimensional vector space $V$ such that $T:Vrightarrow V$, $T$'s injectivity is equivalent to its surjectivity. I started by trying to show $T$'s surjectivity implies its injectivity by
Surjectivity of $T leftrightarrow forall w in V, exists v in V$ s.t. $ Tv = w.$
Let $v = v^ie_i$ for some basis ${e_i}$ of $V$.
$w = v^i(T e_i) = v^ie'_i$.
Surjectivity of $T$ now implies that the ${ e'_i}$ are another (linearly independent) set of basis vectors.
Linear independence of ${e'_i}$ implies that $ineq j rightarrow e_i'-e'_j neq 0$ or $ e_i'-e'_j = 0 rightarrow i = j$ or $Te_i = Te_j rightarrow e_i = e_j leftrightarrow T$ is injective.
Firsly, is this reasoning sound? Secondly, how would I go about showing the opposite statement, that $T$'s injectivity implies its surjectivity?
linear-transformations
linear-transformations
asked Jul 16 '16 at 11:36
user2582713user2582713
136
asked Jul 16 '16 at 11:36
user2582713user2582713
136
edited Jul 16 '16 at 13:02
user2582713
asked Jul 16 '16 at 11:36
user2582713user2582713
136
asked Jul 16 '16 at 11:36
user2582713user2582713
136
asked Jul 16 '16 at 11:36
user2582713user2582713
136
136
$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26
$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27
$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12
$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15
$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18
|
show 7 more comments
$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26
$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27
$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12
$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15
$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18
$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26
$begingroup$
Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
$endgroup$
– user39082
Jul 16 '16 at 12:26
$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27
$begingroup$
For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
$endgroup$
– user39082
Jul 16 '16 at 12:27
$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12
$begingroup$
Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
$endgroup$
– user2582713
Jul 16 '16 at 13:12
$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15
$begingroup$
I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
$endgroup$
– John Coleman
Jul 16 '16 at 13:15
$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18
$begingroup$
@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
$endgroup$
– user39082
Jul 16 '16 at 13:18
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces.
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
$endgroup$
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
add a comment |
$begingroup$
Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.
Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.
Injectivity ===> Surjectivity
The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces.
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
$endgroup$
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
add a comment |
$begingroup$
No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces.
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
$endgroup$
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
add a comment |
$begingroup$
No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces.
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
$endgroup$
No such statement can be true for infinite-dimensional vector spaces. For example, let $V$ be a vector space with a countable basis $left{e_nright}_{nin{mathbb N}}$, then
$$Te_i=e_{i+1} forall iin{mathbb N}$$
defines an injective but not surjective operator, and
$$Te_0=e_0, Te_i=e_{i-1} forall ige 1$$
defines a surjective but not injective operator.
However the equivalence is true for finite-dimensional vector spaces.
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
answered Jul 16 '16 at 11:49
user39082user39082
1,237513
1,237513
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
add a comment |
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
$begingroup$
Thanks for your answer. I forgot to add that V is finite-dimensional. Could you explain why the equivalence is true in this case?
$endgroup$
– user2582713
Jul 16 '16 at 11:58
add a comment |
$begingroup$
Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.
Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.
Injectivity ===> Surjectivity
The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
$endgroup$
add a comment |
$begingroup$
Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.
Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.
Injectivity ===> Surjectivity
The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
$endgroup$
add a comment |
$begingroup$
Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.
Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.
Injectivity ===> Surjectivity
The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
$endgroup$
Surjectivity ===> Injectivity:
Assume by contradiction $T$ is not injective, let $left{v_1,ldots,v_mright}$ be a basis of $ker(T)$ and complete to a basis $left{v_1,ldots,v_nright}$ of $V$. Here $nge mge 1$.
Surjectivity implies that $left{Tv_{m+1},ldots,Tv_nright}$ is a basis of $V$. Indeed any $win V$ is of the form $w=Tv$ and $v$ is of the form $v=a_1v_1+ldots+a_nv_n$, so $w=a_1Tv_1+ldots+a_nTv_n=a_{m+1}Tv_{m+1}+ldots+Tv_n$, so they form a generating system and moreover we know that no linear combination of them can be $0$ because otherwise the corresponding linear combination of $v_i$'s would belong to $ker(T)$.
But this means $dim(V)=n-m$, so $m=0$ and hence $ker(T)=0$.
Injectivity ===> Surjectivity
The argument is similar. For some basis $left{v_1,ldots,v_nright}$ consider the images $left{Tv_1,ldots,Tv_nright}$. From injectivity one gets that they are linearly independent, so because of $dim(V)=n$ they must span all of $V$ and this means surjectivity.
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
answered Jul 16 '16 at 12:23
user39082user39082
1,237513
1,237513
add a comment |
add a comment |
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Surjectivity of $T$ implies linear independence of the $left{e_i^primeright}$ by some dimension count: if they were not linear independent, then one would have a basis with less elements than the original one, which is not possible.
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– user39082
Jul 16 '16 at 12:26
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For the second part, in principle the argument works like this, but of course you have to apply it to any linear combination, not just to differences $e_i-e_j$.
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– user39082
Jul 16 '16 at 12:27
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Do you mean I should impose injectivity of T on a arbitrary linear combination of ${e_i}$? I'm not sure how to proceed.
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– user2582713
Jul 16 '16 at 13:12
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I've always thought of this as a rather immediate corollary of the standard result that if $T:V rightarrow W$ is a linear transformation between finite dimensional vector spaces then dim(V) = dim(ker(T)) + dim(range(T)). If you have already proved this result, there is no reason to reinvent the wheel (and in effect reprove it in a special case)
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– John Coleman
Jul 16 '16 at 13:15
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@John Coleman: this is true, but the proof of both results is essentially the same, so why not think about it in the simpler case?
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– user39082
Jul 16 '16 at 13:18