$n$-dimensional $mathbb Z_2$ vector spaces [on hold]
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Let $v_iinmathbb{Z}_2^n$, $1leq ileq k$ be linearly independent vectors. (defined using $mathbb{Z}_2={0,1}$ field) And let $C$ denote the span.
$$C=langle v_1,v_2,cdots v_krangle$$
Prove that the set of all vectors in $mathbb{Z}_2^n$ orthogonal to $C$, is also a vector space and has dimension $d=n-k$.
P.S. The dot-product is:
$$langle u|vrangleequivsum_{i=1}^n u_i v_ihspace{3mm}{mod2}$$
linear-algebra group-theory coding-theory
edited Nov 20 at 14:12
amWhy
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
put on hold as off-topic by Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz Nov 20 at 14:32
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
Let $v_iinmathbb{Z}_2^n$, $1leq ileq k$ be linearly independent vectors. (defined using $mathbb{Z}_2={0,1}$ field) And let $C$ denote the span.
$$C=langle v_1,v_2,cdots v_krangle$$
Prove that the set of all vectors in $mathbb{Z}_2^n$ orthogonal to $C$, is also a vector space and has dimension $d=n-k$.
P.S. The dot-product is:
$$langle u|vrangleequivsum_{i=1}^n u_i v_ihspace{3mm}{mod2}$$
linear-algebra group-theory coding-theory
edited Nov 20 at 14:12
amWhy
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
put on hold as off-topic by Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz Nov 20 at 14:32
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
1
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
1
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $v_iinmathbb{Z}_2^n$, $1leq ileq k$ be linearly independent vectors. (defined using $mathbb{Z}_2={0,1}$ field) And let $C$ denote the span.
$$C=langle v_1,v_2,cdots v_krangle$$
Prove that the set of all vectors in $mathbb{Z}_2^n$ orthogonal to $C$, is also a vector space and has dimension $d=n-k$.
P.S. The dot-product is:
$$langle u|vrangleequivsum_{i=1}^n u_i v_ihspace{3mm}{mod2}$$
linear-algebra group-theory coding-theory
edited Nov 20 at 14:12
amWhy
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
Let $v_iinmathbb{Z}_2^n$, $1leq ileq k$ be linearly independent vectors. (defined using $mathbb{Z}_2={0,1}$ field) And let $C$ denote the span.
$$C=langle v_1,v_2,cdots v_krangle$$
Prove that the set of all vectors in $mathbb{Z}_2^n$ orthogonal to $C$, is also a vector space and has dimension $d=n-k$.
P.S. The dot-product is:
$$langle u|vrangleequivsum_{i=1}^n u_i v_ihspace{3mm}{mod2}$$
linear-algebra group-theory coding-theory
linear-algebra group-theory coding-theory
edited Nov 20 at 14:12
amWhy
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
edited Nov 20 at 14:12
amWhy
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
edited Nov 20 at 14:12
amWhy
191k27223437
edited Nov 20 at 14:12
amWhy
191k27223437
edited Nov 20 at 14:12
amWhy
191k27223437
191k27223437
asked Nov 20 at 10:25
K. Sadri
656
asked Nov 20 at 10:25
K. Sadri
656
asked Nov 20 at 10:25
K. Sadri
656
656
put on hold as off-topic by Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz Nov 20 at 14:32
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz Nov 20 at 14:32
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Christopher, amWhy, Adrian Keister, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
1
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
1
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20
add a comment |
1
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
1
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
1
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20
1
1
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
1
1
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
1
1
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20
add a comment |
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1
Just for the sake of completeness, as we don't really need it: "orthogonal"....wrt what inner product?
– DonAntonio
Nov 20 at 10:28
1
“Orthogonal” in which sense?
– José Carlos Santos
Nov 20 at 10:28
It is completely straightforward to prove that it is a vector space, so you should be able to at least do that by yourself.
– Derek Holt
Nov 20 at 10:39
Yes, I am able to do that part by myself :))) The second part is tricky however!
– K. Sadri
Nov 20 at 10:45
1
There is nothing special about the field of order $2$ here. The same proof works over any field. You are looking for the nullspace of an $k times n$ matrix of rank $k$, which has rank $n-k$.
– Derek Holt
Nov 20 at 11:20