Mixing
Prove every subspace of a finite-dimensional vector space is finite-dimensional.
$begingroup$
Similar question But don't directly solver my confusions.
In Linear Algebra Done Right, it said:
Proof:
Step 1:
If $U = {0}$, then $U$ is finite-dimensional and we are done. If $U neq {0}$, then choose a nonzero vector $v_1 in U$.
Step J:
If $U = span(v_1,...,v_{j-1})$, then $U$ is finite-dimensional and we are done. If $U neq span(v_1,...,v_{j-1})$, then choose a vector $v_j in U$ such that $v_j not in span(v_1,...,v_{j-1})$.
After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is the span of the previous vectors. Thus, after each step we have constructed a linearly independent list, by Linear Dependence Lemma. This linearly independent list cannot be longer than any spanning list of $V$. Thus the process eventually terminates, which means $U$ is finite-dimensional.
I have problem on "this linearly independent list cannot be longer than any spanning list of $V$." Should it be $U$?
If it is $U$, how do we know the spanning list of $U$ is finite? Or is it the definition, because I remember in Chapter 1, it said list should have finite finite length.
linear-algebra
asked Jan 7 at 22:46
JOHN JOHN
1868
$endgroup$
add a comment |
$begingroup$
Similar question But don't directly solver my confusions.
In Linear Algebra Done Right, it said:
Proof:
Step 1:
If $U = {0}$, then $U$ is finite-dimensional and we are done. If $U neq {0}$, then choose a nonzero vector $v_1 in U$.
Step J:
If $U = span(v_1,...,v_{j-1})$, then $U$ is finite-dimensional and we are done. If $U neq span(v_1,...,v_{j-1})$, then choose a vector $v_j in U$ such that $v_j not in span(v_1,...,v_{j-1})$.
After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is the span of the previous vectors. Thus, after each step we have constructed a linearly independent list, by Linear Dependence Lemma. This linearly independent list cannot be longer than any spanning list of $V$. Thus the process eventually terminates, which means $U$ is finite-dimensional.
I have problem on "this linearly independent list cannot be longer than any spanning list of $V$." Should it be $U$?
If it is $U$, how do we know the spanning list of $U$ is finite? Or is it the definition, because I remember in Chapter 1, it said list should have finite finite length.
linear-algebra
asked Jan 7 at 22:46
JOHN JOHN
1868
$endgroup$
$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53
add a comment |
$begingroup$
Similar question But don't directly solver my confusions.
In Linear Algebra Done Right, it said:
Proof:
Step 1:
If $U = {0}$, then $U$ is finite-dimensional and we are done. If $U neq {0}$, then choose a nonzero vector $v_1 in U$.
Step J:
If $U = span(v_1,...,v_{j-1})$, then $U$ is finite-dimensional and we are done. If $U neq span(v_1,...,v_{j-1})$, then choose a vector $v_j in U$ such that $v_j not in span(v_1,...,v_{j-1})$.
After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is the span of the previous vectors. Thus, after each step we have constructed a linearly independent list, by Linear Dependence Lemma. This linearly independent list cannot be longer than any spanning list of $V$. Thus the process eventually terminates, which means $U$ is finite-dimensional.
I have problem on "this linearly independent list cannot be longer than any spanning list of $V$." Should it be $U$?
If it is $U$, how do we know the spanning list of $U$ is finite? Or is it the definition, because I remember in Chapter 1, it said list should have finite finite length.
linear-algebra
asked Jan 7 at 22:46
JOHN JOHN
1868
$endgroup$
Similar question But don't directly solver my confusions.
In Linear Algebra Done Right, it said:
Proof:
Step 1:
If $U = {0}$, then $U$ is finite-dimensional and we are done. If $U neq {0}$, then choose a nonzero vector $v_1 in U$.
Step J:
If $U = span(v_1,...,v_{j-1})$, then $U$ is finite-dimensional and we are done. If $U neq span(v_1,...,v_{j-1})$, then choose a vector $v_j in U$ such that $v_j not in span(v_1,...,v_{j-1})$.
After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is the span of the previous vectors. Thus, after each step we have constructed a linearly independent list, by Linear Dependence Lemma. This linearly independent list cannot be longer than any spanning list of $V$. Thus the process eventually terminates, which means $U$ is finite-dimensional.
I have problem on "this linearly independent list cannot be longer than any spanning list of $V$." Should it be $U$?
If it is $U$, how do we know the spanning list of $U$ is finite? Or is it the definition, because I remember in Chapter 1, it said list should have finite finite length.
linear-algebra
linear-algebra
asked Jan 7 at 22:46
JOHN JOHN
1868
asked Jan 7 at 22:46
JOHN JOHN
1868
asked Jan 7 at 22:46
JOHN JOHN
1868
asked Jan 7 at 22:46
JOHN JOHN
1868
asked Jan 7 at 22:46
JOHN JOHN
1868
1868
$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53
add a comment |
$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53
$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53
add a comment |
2 Answers
2
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$begingroup$
It is supposed to be $V$. The proof constructs linearly independent $v_1, v_2, ldots$, which are going to be linearly independent regardless of whether you consider them in $U$ or $V$.
Because they're linearly independent as vectors in $V$, the process must terminate eventually. It won't always terminate after $operatorname{dim} V$ vectors (that would imply $U = V$), but there is always this fixed upper bound $operatorname{dim} V$ that is independent of $U$. That should answer your second question.
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
$endgroup$
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
add a comment |
$begingroup$
No, it should be $V$ as written.
The point is that the constructed sequence $v_1,v_2,dots$ lies in $V$, and is linearly independent by construction.
answered Jan 7 at 22:53
BerciBerci
60.3k23672
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is supposed to be $V$. The proof constructs linearly independent $v_1, v_2, ldots$, which are going to be linearly independent regardless of whether you consider them in $U$ or $V$.
Because they're linearly independent as vectors in $V$, the process must terminate eventually. It won't always terminate after $operatorname{dim} V$ vectors (that would imply $U = V$), but there is always this fixed upper bound $operatorname{dim} V$ that is independent of $U$. That should answer your second question.
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
$endgroup$
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
add a comment |
$begingroup$
It is supposed to be $V$. The proof constructs linearly independent $v_1, v_2, ldots$, which are going to be linearly independent regardless of whether you consider them in $U$ or $V$.
Because they're linearly independent as vectors in $V$, the process must terminate eventually. It won't always terminate after $operatorname{dim} V$ vectors (that would imply $U = V$), but there is always this fixed upper bound $operatorname{dim} V$ that is independent of $U$. That should answer your second question.
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
$endgroup$
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
add a comment |
$begingroup$
It is supposed to be $V$. The proof constructs linearly independent $v_1, v_2, ldots$, which are going to be linearly independent regardless of whether you consider them in $U$ or $V$.
Because they're linearly independent as vectors in $V$, the process must terminate eventually. It won't always terminate after $operatorname{dim} V$ vectors (that would imply $U = V$), but there is always this fixed upper bound $operatorname{dim} V$ that is independent of $U$. That should answer your second question.
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
$endgroup$
It is supposed to be $V$. The proof constructs linearly independent $v_1, v_2, ldots$, which are going to be linearly independent regardless of whether you consider them in $U$ or $V$.
Because they're linearly independent as vectors in $V$, the process must terminate eventually. It won't always terminate after $operatorname{dim} V$ vectors (that would imply $U = V$), but there is always this fixed upper bound $operatorname{dim} V$ that is independent of $U$. That should answer your second question.
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
answered Jan 7 at 23:10
Theo BenditTheo Bendit
17.4k12150
17.4k12150
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
add a comment |
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate.
$endgroup$
– JOHN
Jan 9 at 23:33
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
$begingroup$
@JOHN Yes, precisely.
$endgroup$
– Theo Bendit
Jan 10 at 1:24
add a comment |
$begingroup$
No, it should be $V$ as written.
The point is that the constructed sequence $v_1,v_2,dots$ lies in $V$, and is linearly independent by construction.
answered Jan 7 at 22:53
BerciBerci
60.3k23672
$endgroup$
add a comment |
$begingroup$
No, it should be $V$ as written.
The point is that the constructed sequence $v_1,v_2,dots$ lies in $V$, and is linearly independent by construction.
answered Jan 7 at 22:53
BerciBerci
60.3k23672
$endgroup$
add a comment |
$begingroup$
No, it should be $V$ as written.
The point is that the constructed sequence $v_1,v_2,dots$ lies in $V$, and is linearly independent by construction.
answered Jan 7 at 22:53
BerciBerci
60.3k23672
$endgroup$
No, it should be $V$ as written.
The point is that the constructed sequence $v_1,v_2,dots$ lies in $V$, and is linearly independent by construction.
answered Jan 7 at 22:53
BerciBerci
60.3k23672
answered Jan 7 at 22:53
BerciBerci
60.3k23672
answered Jan 7 at 22:53
BerciBerci
60.3k23672
answered Jan 7 at 22:53
BerciBerci
60.3k23672
60.3k23672
add a comment |
add a comment |
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$begingroup$
You can replace the word "list" by "set" if that clarifies things.
$endgroup$
– Math_QED
Jan 7 at 23:33
$begingroup$
Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $dim Ule dim V<infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ...
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:50
$begingroup$
@Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation.
$endgroup$
– Hagen von Eitzen
Jan 8 at 11:53