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Prove: $dim E= +infty Longleftrightarrow forall ;ngeq 1,;exists ;;B=u_1,u_2,cdots,u_n$ such that $B$ is Linearly independent
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I wish to prove that $$dim E= +infty Longleftrightarrow forall ;ngeq 1,;exists ;u_1,u_2,cdots,u_n; textsuch that; u_1,u_2,cdots,u_n ;;text is linearly independent.$$This looks completely new to me, so I would love if someone could give me an elaborate answer or provide a reference.
Thanks!
linear-algebra functional-analysis analysis
asked Jul 27 at 17:02
Mike
62112
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1
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I wish to prove that $$dim E= +infty Longleftrightarrow forall ;ngeq 1,;exists ;u_1,u_2,cdots,u_n; textsuch that; u_1,u_2,cdots,u_n ;;text is linearly independent.$$This looks completely new to me, so I would love if someone could give me an elaborate answer or provide a reference.
Thanks!
linear-algebra functional-analysis analysis
asked Jul 27 at 17:02
Mike
62112
Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I wish to prove that $$dim E= +infty Longleftrightarrow forall ;ngeq 1,;exists ;u_1,u_2,cdots,u_n; textsuch that; u_1,u_2,cdots,u_n ;;text is linearly independent.$$This looks completely new to me, so I would love if someone could give me an elaborate answer or provide a reference.
Thanks!
linear-algebra functional-analysis analysis
asked Jul 27 at 17:02
Mike
62112
I wish to prove that $$dim E= +infty Longleftrightarrow forall ;ngeq 1,;exists ;u_1,u_2,cdots,u_n; textsuch that; u_1,u_2,cdots,u_n ;;text is linearly independent.$$This looks completely new to me, so I would love if someone could give me an elaborate answer or provide a reference.
Thanks!
linear-algebra functional-analysis analysis
asked Jul 27 at 17:02
Mike
62112
edited Jul 27 at 18:30
asked Jul 27 at 17:02
Mike
62112
asked Jul 27 at 17:02
Mike
62112
asked Jul 27 at 17:02
Mike
62112
62112
Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29
add a comment |Â
Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29
Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29
add a comment |Â
1 Answer
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Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:
For right to left: Suppose that $mathrmdim(E)<+infty$, i.e. $mathrmdim(E)=m$ f.s. $minmathbbN$. Then For all $n>m$, any set $u_1,dots,u_n$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)
For left to right: Suppose that $mathrmdim(E)=+infty$. Suppose that there exists a $ngeq 1$ such that for all sets $U=u_1,dots,u_n$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $UcupW$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite)
Now, as $E$ is infinite dimensional, we have a basis $B=b_imid iin I$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
answered Jul 27 at 18:14
zzuussee
1,442419
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:
For right to left: Suppose that $mathrmdim(E)<+infty$, i.e. $mathrmdim(E)=m$ f.s. $minmathbbN$. Then For all $n>m$, any set $u_1,dots,u_n$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)
For left to right: Suppose that $mathrmdim(E)=+infty$. Suppose that there exists a $ngeq 1$ such that for all sets $U=u_1,dots,u_n$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $UcupW$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite)
Now, as $E$ is infinite dimensional, we have a basis $B=b_imid iin I$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
answered Jul 27 at 18:14
zzuussee
1,442419
add a comment |Â
up vote
1
down vote
accepted
Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:
For right to left: Suppose that $mathrmdim(E)<+infty$, i.e. $mathrmdim(E)=m$ f.s. $minmathbbN$. Then For all $n>m$, any set $u_1,dots,u_n$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)
For left to right: Suppose that $mathrmdim(E)=+infty$. Suppose that there exists a $ngeq 1$ such that for all sets $U=u_1,dots,u_n$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $UcupW$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite)
Now, as $E$ is infinite dimensional, we have a basis $B=b_imid iin I$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
answered Jul 27 at 18:14
zzuussee
1,442419
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:
For right to left: Suppose that $mathrmdim(E)<+infty$, i.e. $mathrmdim(E)=m$ f.s. $minmathbbN$. Then For all $n>m$, any set $u_1,dots,u_n$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)
For left to right: Suppose that $mathrmdim(E)=+infty$. Suppose that there exists a $ngeq 1$ such that for all sets $U=u_1,dots,u_n$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $UcupW$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite)
Now, as $E$ is infinite dimensional, we have a basis $B=b_imid iin I$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
answered Jul 27 at 18:14
zzuussee
1,442419
Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:
For right to left: Suppose that $mathrmdim(E)<+infty$, i.e. $mathrmdim(E)=m$ f.s. $minmathbbN$. Then For all $n>m$, any set $u_1,dots,u_n$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)
For left to right: Suppose that $mathrmdim(E)=+infty$. Suppose that there exists a $ngeq 1$ such that for all sets $U=u_1,dots,u_n$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $UcupW$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite)
Now, as $E$ is infinite dimensional, we have a basis $B=b_imid iin I$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
answered Jul 27 at 18:14
zzuussee
1,442419
edited Jul 27 at 18:24
answered Jul 27 at 18:14
zzuussee
1,442419
answered Jul 27 at 18:14
zzuussee
1,442419
answered Jul 27 at 18:14
zzuussee
1,442419
1,442419
add a comment |Â
add a comment |Â
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Nice example could be polynomials
– Ninja hatori
Jul 27 at 17:06
What's the purpose of $f$ in this question? Also, this question would be greatly improved if you include some of your own thoughts and context: What is your definition of dimension, and of infinite dimension? The problem statement is an if and only if: Can you see any links between the incomplete basis theorem or definition of dimension and either direction of the question?
– B. Mehta
Jul 27 at 17:20
@B. Mehta: Sorry for that! I would improve the question!
– Mike
Jul 27 at 18:29