Mixing
suppose that $V$ is a finite dimensional and $U$ is a subspace of $V$ such that $dim U = dim V$. Prove $U =...
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I get stuck in this exercise. Can some one give me a hand? It is from Linear Algebra Done Right, from Sheldon Axler.
linear-algebra
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amWhy
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Jeremias Junior
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I get stuck in this exercise. Can some one give me a hand? It is from Linear Algebra Done Right, from Sheldon Axler.
linear-algebra
edited yesterday
amWhy
191k27223437
asked yesterday
Jeremias Junior
1
New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I get stuck in this exercise. Can some one give me a hand? It is from Linear Algebra Done Right, from Sheldon Axler.
linear-algebra
edited yesterday
amWhy
191k27223437
asked yesterday
Jeremias Junior
1
New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I get stuck in this exercise. Can some one give me a hand? It is from Linear Algebra Done Right, from Sheldon Axler.
linear-algebra
linear-algebra
edited yesterday
amWhy
191k27223437
asked yesterday
Jeremias Junior
1
New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
amWhy
191k27223437
asked yesterday
Jeremias Junior
1
New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
191k27223437
asked yesterday
Jeremias Junior
1
New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
Jeremias Junior
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asked yesterday
Jeremias Junior
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Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jeremias Junior is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday
|
show 3 more comments
1
Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday
1
1
Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday
|
show 3 more comments
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Consider a basis of $U$. Perhaps it is also a basis of $V$.
– ajotatxe
yesterday
What do you know about the concept of "dimension" in linear algebra? List a couple of properties, and see if you can't find one which you can use as leverage to your advantage.
– Arthur
yesterday
Ajotaxe, i did it, in my proof the basis of V must be equal to the basis of U, to V=U. But i really dont know if is what the exercice propose. I assume that existe v belonging to V that must belong to U as well. And i open up everything with a linear combination. then i did v-v, and again open up as a linear combination. I got b1.u1 + ... + bn.un = a1v1 + ... + anvn. But i really think this is wrong, or not a good proof.
– Jeremias Junior
yesterday
Arthur, i will try again. And see if i miss something.
– Jeremias Junior
yesterday
@ajotatxe if U is a subspace of V and in this case dim U = dim V, then the basis of U must be the same of V, by definition ?
– Jeremias Junior
yesterday