Let A and B be two sets of sequences. I want to say whether or not they are orthonormal basis for $l^2$
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Let A and B be two sets of sequences. I want to say whether or not they are orthonormal basis for $l^2$. More specific:
A = { (1,2,0,0,0,0...), (0,0,1,2,0,0,0,0,..), (0,0,0,0,1,2,0,0,0,0,..) }
B = { (1,-1,0,0,0,0,...), (1,1,0,0,0,0,...), (0,0,1,-1,0,0,0,0), (0,0,1,1,0,0,0,0)}
Now, I know that a set is said to be orthonormal basis of H if it is an orthonormal set and it is complete. Both A and B are orthonormal, but I am struggling to see if they are complete. To disprove this fact I would need to find a sequence in $l^2$ such that it is not made by elements of A, but I am also struggling to see how to prove this statement in positive.
EDIT: For A), I think that (1,0,0,0,0,0...) is in $l^2$ but cannot be made by elements in A, is that right?
functional-analysis
asked Nov 28 at 15:40
qcc101
458113
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up vote
0
down vote
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Let A and B be two sets of sequences. I want to say whether or not they are orthonormal basis for $l^2$. More specific:
A = { (1,2,0,0,0,0...), (0,0,1,2,0,0,0,0,..), (0,0,0,0,1,2,0,0,0,0,..) }
B = { (1,-1,0,0,0,0,...), (1,1,0,0,0,0,...), (0,0,1,-1,0,0,0,0), (0,0,1,1,0,0,0,0)}
Now, I know that a set is said to be orthonormal basis of H if it is an orthonormal set and it is complete. Both A and B are orthonormal, but I am struggling to see if they are complete. To disprove this fact I would need to find a sequence in $l^2$ such that it is not made by elements of A, but I am also struggling to see how to prove this statement in positive.
EDIT: For A), I think that (1,0,0,0,0,0...) is in $l^2$ but cannot be made by elements in A, is that right?
functional-analysis
asked Nov 28 at 15:40
qcc101
458113
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let A and B be two sets of sequences. I want to say whether or not they are orthonormal basis for $l^2$. More specific:
A = { (1,2,0,0,0,0...), (0,0,1,2,0,0,0,0,..), (0,0,0,0,1,2,0,0,0,0,..) }
B = { (1,-1,0,0,0,0,...), (1,1,0,0,0,0,...), (0,0,1,-1,0,0,0,0), (0,0,1,1,0,0,0,0)}
Now, I know that a set is said to be orthonormal basis of H if it is an orthonormal set and it is complete. Both A and B are orthonormal, but I am struggling to see if they are complete. To disprove this fact I would need to find a sequence in $l^2$ such that it is not made by elements of A, but I am also struggling to see how to prove this statement in positive.
EDIT: For A), I think that (1,0,0,0,0,0...) is in $l^2$ but cannot be made by elements in A, is that right?
functional-analysis
asked Nov 28 at 15:40
qcc101
458113
Let A and B be two sets of sequences. I want to say whether or not they are orthonormal basis for $l^2$. More specific:
A = { (1,2,0,0,0,0...), (0,0,1,2,0,0,0,0,..), (0,0,0,0,1,2,0,0,0,0,..) }
B = { (1,-1,0,0,0,0,...), (1,1,0,0,0,0,...), (0,0,1,-1,0,0,0,0), (0,0,1,1,0,0,0,0)}
Now, I know that a set is said to be orthonormal basis of H if it is an orthonormal set and it is complete. Both A and B are orthonormal, but I am struggling to see if they are complete. To disprove this fact I would need to find a sequence in $l^2$ such that it is not made by elements of A, but I am also struggling to see how to prove this statement in positive.
EDIT: For A), I think that (1,0,0,0,0,0...) is in $l^2$ but cannot be made by elements in A, is that right?
functional-analysis
functional-analysis
asked Nov 28 at 15:40
qcc101
458113
asked Nov 28 at 15:40
qcc101
458113
asked Nov 28 at 15:40
qcc101
458113
asked Nov 28 at 15:40
qcc101
458113
asked Nov 28 at 15:40
qcc101
458113
458113
add a comment |
add a comment |
1 Answer
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$(-2,1,0,0,0,cdots)$ is orthogonal to everything in $A$. So $A$ is not an orthogonal basis.
The second set is an orthogonal basis; all the standard basis elements can be written in terms of these elements, and they are mutually orthogonal.
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$(-2,1,0,0,0,cdots)$ is orthogonal to everything in $A$. So $A$ is not an orthogonal basis.
The second set is an orthogonal basis; all the standard basis elements can be written in terms of these elements, and they are mutually orthogonal.
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
add a comment |
up vote
0
down vote
accepted
$(-2,1,0,0,0,cdots)$ is orthogonal to everything in $A$. So $A$ is not an orthogonal basis.
The second set is an orthogonal basis; all the standard basis elements can be written in terms of these elements, and they are mutually orthogonal.
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$(-2,1,0,0,0,cdots)$ is orthogonal to everything in $A$. So $A$ is not an orthogonal basis.
The second set is an orthogonal basis; all the standard basis elements can be written in terms of these elements, and they are mutually orthogonal.
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
$(-2,1,0,0,0,cdots)$ is orthogonal to everything in $A$. So $A$ is not an orthogonal basis.
The second set is an orthogonal basis; all the standard basis elements can be written in terms of these elements, and they are mutually orthogonal.
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
answered Nov 28 at 19:45
DisintegratingByParts
58.3k42579
58.3k42579
add a comment |
add a comment |
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