Mixing
$ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$ where $R$ is an isomorphism?
1
$begingroup$
Let $V$ be a finite-dimensional vector space, and let $R: V rightarrow V$ be an invertible linear transformation.
Is there a linear transformation $T: V rightarrow V$ such that $ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$?
What I'd like you to help me with:
- As I understand it (and my understanding is probably wrong), $R$ defines an isomorphism between $V$ to $V$ itself; But any vector space is isomorphic with itself, so it seems like a trivial thing...
- I need some clues on how to approach the proof. I guess it should involve using dimensions, but I don't know how to start applying the relevant dimensions theorems
linear-algebra linear-transformations vector-space-isomorphism
edited Jan 13 at 23:55
Bernard
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
$endgroup$
add a comment |
1
$begingroup$
Let $V$ be a finite-dimensional vector space, and let $R: V rightarrow V$ be an invertible linear transformation.
Is there a linear transformation $T: V rightarrow V$ such that $ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$?
What I'd like you to help me with:
- As I understand it (and my understanding is probably wrong), $R$ defines an isomorphism between $V$ to $V$ itself; But any vector space is isomorphic with itself, so it seems like a trivial thing...
- I need some clues on how to approach the proof. I guess it should involve using dimensions, but I don't know how to start applying the relevant dimensions theorems
linear-algebra linear-transformations vector-space-isomorphism
edited Jan 13 at 23:55
Bernard
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
$endgroup$
add a comment |
1
1
1
$begingroup$
Let $V$ be a finite-dimensional vector space, and let $R: V rightarrow V$ be an invertible linear transformation.
Is there a linear transformation $T: V rightarrow V$ such that $ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$?
What I'd like you to help me with:
- As I understand it (and my understanding is probably wrong), $R$ defines an isomorphism between $V$ to $V$ itself; But any vector space is isomorphic with itself, so it seems like a trivial thing...
- I need some clues on how to approach the proof. I guess it should involve using dimensions, but I don't know how to start applying the relevant dimensions theorems
linear-algebra linear-transformations vector-space-isomorphism
edited Jan 13 at 23:55
Bernard
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
$endgroup$
Let $V$ be a finite-dimensional vector space, and let $R: V rightarrow V$ be an invertible linear transformation.
Is there a linear transformation $T: V rightarrow V$ such that $ operatorname{Ker}TR = {0}$ but $ operatorname{Ker}T not= {0}$?
What I'd like you to help me with:
- As I understand it (and my understanding is probably wrong), $R$ defines an isomorphism between $V$ to $V$ itself; But any vector space is isomorphic with itself, so it seems like a trivial thing...
- I need some clues on how to approach the proof. I guess it should involve using dimensions, but I don't know how to start applying the relevant dimensions theorems
linear-algebra linear-transformations vector-space-isomorphism
linear-algebra linear-transformations vector-space-isomorphism
edited Jan 13 at 23:55
Bernard
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
edited Jan 13 at 23:55
Bernard
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
edited Jan 13 at 23:55
Bernard
121k740116
edited Jan 13 at 23:55
Bernard
121k740116
edited Jan 13 at 23:55
Bernard
121k740116
121k740116
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
asked Jan 13 at 23:47
HeyJudeHeyJude
1687
1687
add a comment |
add a comment |
1 Answer
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oldest
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3
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It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
add a comment |
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1 Answer
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1 Answer
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3
$begingroup$
It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
add a comment |
3
$begingroup$
It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
add a comment |
3
3
3
$begingroup$
It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
$endgroup$
It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
answered Jan 13 at 23:51
Tsemo AristideTsemo Aristide
58.1k11445
58.1k11445
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
add a comment |
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Worth noting this uses the invertibility of $R$ (specifically that it is surjective); were it not surjective this would not be true
$endgroup$
– aidangallagher4
Jan 14 at 0:07
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can you please explain how do we use the kernels here to finish the proof?
$endgroup$
– HeyJude
Jan 14 at 0:58
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
$begingroup$
Can we just say that $KerT$ must be equal to $KerTR$?
$endgroup$
– HeyJude
Jan 14 at 1:56
add a comment |
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