Mixing
Orthogonal bipolar vectors in 3-dimensions [closed]
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Suppose the vectors must consist of only 1 and - 1 (bipolar). This binary encoding is common for many neural networks.
Then can two 3D bipolar vectors be orthogonal? Orthogonal vectors can be successfully stored and retrieved in an auto-association memory network (like Hopfield).
As I try to figure out a dot product of any two of such vectors to be zero, I can't find one.
linear-algebra
asked Jan 13 at 15:58
AhmadAhmad
22418
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closed as off-topic by Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn Jan 14 at 7:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose the vectors must consist of only 1 and - 1 (bipolar). This binary encoding is common for many neural networks.
Then can two 3D bipolar vectors be orthogonal? Orthogonal vectors can be successfully stored and retrieved in an auto-association memory network (like Hopfield).
As I try to figure out a dot product of any two of such vectors to be zero, I can't find one.
linear-algebra
asked Jan 13 at 15:58
AhmadAhmad
22418
$endgroup$
closed as off-topic by Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn Jan 14 at 7:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
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– M. Winter
Jan 19 at 12:39
add a comment |
$begingroup$
Suppose the vectors must consist of only 1 and - 1 (bipolar). This binary encoding is common for many neural networks.
Then can two 3D bipolar vectors be orthogonal? Orthogonal vectors can be successfully stored and retrieved in an auto-association memory network (like Hopfield).
As I try to figure out a dot product of any two of such vectors to be zero, I can't find one.
linear-algebra
asked Jan 13 at 15:58
AhmadAhmad
22418
$endgroup$
Suppose the vectors must consist of only 1 and - 1 (bipolar). This binary encoding is common for many neural networks.
Then can two 3D bipolar vectors be orthogonal? Orthogonal vectors can be successfully stored and retrieved in an auto-association memory network (like Hopfield).
As I try to figure out a dot product of any two of such vectors to be zero, I can't find one.
linear-algebra
linear-algebra
asked Jan 13 at 15:58
AhmadAhmad
22418
asked Jan 13 at 15:58
AhmadAhmad
22418
edited Jan 19 at 12:45
Ahmad
asked Jan 13 at 15:58
AhmadAhmad
22418
asked Jan 13 at 15:58
AhmadAhmad
22418
asked Jan 13 at 15:58
AhmadAhmad
22418
22418
closed as off-topic by Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn Jan 14 at 7:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn Jan 14 at 7:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, RRL, Lord Shark the Unknown, mrtaurho, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
$endgroup$
– M. Winter
Jan 19 at 12:39
add a comment |
2
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
$endgroup$
– M. Winter
Jan 19 at 12:39
2
2
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
$endgroup$
– M. Winter
Jan 19 at 12:39
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
$endgroup$
– M. Winter
Jan 19 at 12:39
add a comment |
1 Answer
1
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There are only eight vectors in three dimensions whose entries are $pm1$. The summands of the dot product between any two of these vectors are also $pm1$, so if the dot product were to be zero there would have to be equal numbers of $+1$ and $-1$ summands, hence an even-length vector. But 3 is an odd number, so no two 3-dimensional "bipolar" vectors are orthogonal.
The same result applies to other odd-dimensional Euclidean vector spaces.
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are only eight vectors in three dimensions whose entries are $pm1$. The summands of the dot product between any two of these vectors are also $pm1$, so if the dot product were to be zero there would have to be equal numbers of $+1$ and $-1$ summands, hence an even-length vector. But 3 is an odd number, so no two 3-dimensional "bipolar" vectors are orthogonal.
The same result applies to other odd-dimensional Euclidean vector spaces.
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
There are only eight vectors in three dimensions whose entries are $pm1$. The summands of the dot product between any two of these vectors are also $pm1$, so if the dot product were to be zero there would have to be equal numbers of $+1$ and $-1$ summands, hence an even-length vector. But 3 is an odd number, so no two 3-dimensional "bipolar" vectors are orthogonal.
The same result applies to other odd-dimensional Euclidean vector spaces.
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
There are only eight vectors in three dimensions whose entries are $pm1$. The summands of the dot product between any two of these vectors are also $pm1$, so if the dot product were to be zero there would have to be equal numbers of $+1$ and $-1$ summands, hence an even-length vector. But 3 is an odd number, so no two 3-dimensional "bipolar" vectors are orthogonal.
The same result applies to other odd-dimensional Euclidean vector spaces.
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
There are only eight vectors in three dimensions whose entries are $pm1$. The summands of the dot product between any two of these vectors are also $pm1$, so if the dot product were to be zero there would have to be equal numbers of $+1$ and $-1$ summands, hence an even-length vector. But 3 is an odd number, so no two 3-dimensional "bipolar" vectors are orthogonal.
The same result applies to other odd-dimensional Euclidean vector spaces.
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:01
Parcly TaxelParcly Taxel
41.8k1372101
41.8k1372101
add a comment |
add a comment |
2
$begingroup$
I don't quite get why this question was closed. However, the answer is simple: two such vectors $v,win{-1,1}^{2n}$ (in even dimensions) are orthogonal if and only if they differ in exactly $n$ components. In odd dimensions, no two such vectors can be orthogonal.
$endgroup$
– M. Winter
Jan 19 at 12:39