Mixing
Find basis of $ U = left{ u in mathbb C^n |u_1+u_2 +…+u_n=0 right}$
$begingroup$
I want to find basis of
$$ U = left{ u in mathbb C^n |u_1+u_2 +...+u_n=0 right}$$
I am writing this post because I am not sure with one thing:
If I had :
$$ Z = left{ z in mathbb R^n |z_1+z_2 +...+z_n=0 right}$$
then $$ Z=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
Ok, but If I consider complex space $mathbb C^n$, should I add vector to basis for complex unit?
I mean instead of
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
put
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,[-i,0,i,0,...,0]^T,...,[-i,0,...,0,i^T])$$
linear-algebra complex-analysis
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
I want to find basis of
$$ U = left{ u in mathbb C^n |u_1+u_2 +...+u_n=0 right}$$
I am writing this post because I am not sure with one thing:
If I had :
$$ Z = left{ z in mathbb R^n |z_1+z_2 +...+z_n=0 right}$$
then $$ Z=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
Ok, but If I consider complex space $mathbb C^n$, should I add vector to basis for complex unit?
I mean instead of
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
put
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,[-i,0,i,0,...,0]^T,...,[-i,0,...,0,i^T])$$
linear-algebra complex-analysis
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
I want to find basis of
$$ U = left{ u in mathbb C^n |u_1+u_2 +...+u_n=0 right}$$
I am writing this post because I am not sure with one thing:
If I had :
$$ Z = left{ z in mathbb R^n |z_1+z_2 +...+z_n=0 right}$$
then $$ Z=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
Ok, but If I consider complex space $mathbb C^n$, should I add vector to basis for complex unit?
I mean instead of
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
put
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,[-i,0,i,0,...,0]^T,...,[-i,0,...,0,i^T])$$
linear-algebra complex-analysis
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
$endgroup$
I want to find basis of
$$ U = left{ u in mathbb C^n |u_1+u_2 +...+u_n=0 right}$$
I am writing this post because I am not sure with one thing:
If I had :
$$ Z = left{ z in mathbb R^n |z_1+z_2 +...+z_n=0 right}$$
then $$ Z=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
Ok, but If I consider complex space $mathbb C^n$, should I add vector to basis for complex unit?
I mean instead of
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T])$$
put
$$ U=span([-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,[-i,0,i,0,...,0]^T,...,[-i,0,...,0,i^T])$$
linear-algebra complex-analysis
linear-algebra complex-analysis
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
edited Feb 1 at 21:54
VirtualUser
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
asked Feb 1 at 21:31
VirtualUserVirtualUser
1,321317
1,321317
add a comment |
add a comment |
1 Answer
1
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$begingroup$
In order to define a vector space, you have to specify the set of vectors, the field of scalars, the addition operation, and the scaling operation. In this context, there are at least two ways to view $U$ as a vector space; you could say the field of scalars is $mathbb C$ or $mathbb R$ (and that the addition/scaling operations are the usual ones).
If the field of scalars is $mathbb C$, then
$$
{[-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T]}
$$
is a basis, since every element of $U$ can be uniquely written in the form $sum_{k=1}^{n-1}c_kv_k$ with $v_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb C$.
If the field of scalars is $mathbb R$, then
$$
{[-1,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,...,[-i,0,...,0,i^T]}
$$
is a basis, since every element of $U$ can be written uniquely in the form $sum_{k=1}^{2n-2}c_kw_k$, with $w_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb R$.
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
$endgroup$
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
In order to define a vector space, you have to specify the set of vectors, the field of scalars, the addition operation, and the scaling operation. In this context, there are at least two ways to view $U$ as a vector space; you could say the field of scalars is $mathbb C$ or $mathbb R$ (and that the addition/scaling operations are the usual ones).
If the field of scalars is $mathbb C$, then
$$
{[-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T]}
$$
is a basis, since every element of $U$ can be uniquely written in the form $sum_{k=1}^{n-1}c_kv_k$ with $v_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb C$.
If the field of scalars is $mathbb R$, then
$$
{[-1,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,...,[-i,0,...,0,i^T]}
$$
is a basis, since every element of $U$ can be written uniquely in the form $sum_{k=1}^{2n-2}c_kw_k$, with $w_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb R$.
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
$endgroup$
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
add a comment |
$begingroup$
In order to define a vector space, you have to specify the set of vectors, the field of scalars, the addition operation, and the scaling operation. In this context, there are at least two ways to view $U$ as a vector space; you could say the field of scalars is $mathbb C$ or $mathbb R$ (and that the addition/scaling operations are the usual ones).
If the field of scalars is $mathbb C$, then
$$
{[-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T]}
$$
is a basis, since every element of $U$ can be uniquely written in the form $sum_{k=1}^{n-1}c_kv_k$ with $v_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb C$.
If the field of scalars is $mathbb R$, then
$$
{[-1,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,...,[-i,0,...,0,i^T]}
$$
is a basis, since every element of $U$ can be written uniquely in the form $sum_{k=1}^{2n-2}c_kw_k$, with $w_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb R$.
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
$endgroup$
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
add a comment |
$begingroup$
In order to define a vector space, you have to specify the set of vectors, the field of scalars, the addition operation, and the scaling operation. In this context, there are at least two ways to view $U$ as a vector space; you could say the field of scalars is $mathbb C$ or $mathbb R$ (and that the addition/scaling operations are the usual ones).
If the field of scalars is $mathbb C$, then
$$
{[-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T]}
$$
is a basis, since every element of $U$ can be uniquely written in the form $sum_{k=1}^{n-1}c_kv_k$ with $v_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb C$.
If the field of scalars is $mathbb R$, then
$$
{[-1,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,...,[-i,0,...,0,i^T]}
$$
is a basis, since every element of $U$ can be written uniquely in the form $sum_{k=1}^{2n-2}c_kw_k$, with $w_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb R$.
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
$endgroup$
In order to define a vector space, you have to specify the set of vectors, the field of scalars, the addition operation, and the scaling operation. In this context, there are at least two ways to view $U$ as a vector space; you could say the field of scalars is $mathbb C$ or $mathbb R$ (and that the addition/scaling operations are the usual ones).
If the field of scalars is $mathbb C$, then
$$
{[-1,1,0,...,0]^T,[-1,0,1,0,...,0]^T,...,[-1,0,...,0,1^T]}
$$
is a basis, since every element of $U$ can be uniquely written in the form $sum_{k=1}^{n-1}c_kv_k$ with $v_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb C$.
If the field of scalars is $mathbb R$, then
$$
{[-1,1,0,...,0]^T,...,[-1,0,...,0,1^T],[-i,i,0,...,0]^T,...,[-i,0,...,0,i^T]}
$$
is a basis, since every element of $U$ can be written uniquely in the form $sum_{k=1}^{2n-2}c_kw_k$, with $w_k$ being the $k^{th}$ vector in the above list and $c_kin mathbb R$.
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
edited Feb 1 at 22:06
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
answered Feb 1 at 21:52
Mike EarnestMike Earnest
27.6k22152
27.6k22152
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
add a comment |
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
$begingroup$
Ahh, so the background of scalars is important. Thanks for great explanation!
$endgroup$
– VirtualUser
Feb 1 at 22:43
add a comment |
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