Mixing
Proving generating system for $U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4=:V,Vmathbb{R}$-vectorspace
$begingroup$
I have already proved it if V would be a $mathbb{C}$ vectorspace. I picked $(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$ as my base showed linear Independence and said if there would exist one more basevector the Dimension would be 4 which cannot be because then it would be $mathbb{C}$, which would mean $(1,0,0,0)$ would also be in $U$ which is false. I cannot make the same Argument with $mathbb{R}$, my corresponding base would be $(i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$
of Dimension 6. The Base for V would be of Dimension $8$. how can I now prove that my new base is a generating System?
linear-algebra
asked Jan 14 at 20:03
RM777RM777
39112
$endgroup$
add a comment |
$begingroup$
I have already proved it if V would be a $mathbb{C}$ vectorspace. I picked $(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$ as my base showed linear Independence and said if there would exist one more basevector the Dimension would be 4 which cannot be because then it would be $mathbb{C}$, which would mean $(1,0,0,0)$ would also be in $U$ which is false. I cannot make the same Argument with $mathbb{R}$, my corresponding base would be $(i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$
of Dimension 6. The Base for V would be of Dimension $8$. how can I now prove that my new base is a generating System?
linear-algebra
asked Jan 14 at 20:03
RM777RM777
39112
$endgroup$
1
$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
1
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46
add a comment |
$begingroup$
I have already proved it if V would be a $mathbb{C}$ vectorspace. I picked $(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$ as my base showed linear Independence and said if there would exist one more basevector the Dimension would be 4 which cannot be because then it would be $mathbb{C}$, which would mean $(1,0,0,0)$ would also be in $U$ which is false. I cannot make the same Argument with $mathbb{R}$, my corresponding base would be $(i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$
of Dimension 6. The Base for V would be of Dimension $8$. how can I now prove that my new base is a generating System?
linear-algebra
asked Jan 14 at 20:03
RM777RM777
39112
$endgroup$
I have already proved it if V would be a $mathbb{C}$ vectorspace. I picked $(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$ as my base showed linear Independence and said if there would exist one more basevector the Dimension would be 4 which cannot be because then it would be $mathbb{C}$, which would mean $(1,0,0,0)$ would also be in $U$ which is false. I cannot make the same Argument with $mathbb{R}$, my corresponding base would be $(i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$
of Dimension 6. The Base for V would be of Dimension $8$. how can I now prove that my new base is a generating System?
linear-algebra
linear-algebra
asked Jan 14 at 20:03
RM777RM777
39112
asked Jan 14 at 20:03
RM777RM777
39112
asked Jan 14 at 20:03
RM777RM777
39112
asked Jan 14 at 20:03
RM777RM777
39112
asked Jan 14 at 20:03
RM777RM777
39112
39112
1
$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
1
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46
add a comment |
1
$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
1
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46
1
1
$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
1
1
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the same way as you already did when $mathbb{C}^4$ was considered as $mathbb{C}$-vectorspace and as indicated in the comment by @Did, we show that
begin{align*}
G={&(1,-1,0,0),(0,1,-1,0),(0,0,1,-1),\
&(i,-i,0,0),(0,i,-i,0),(0,0,i,-i)}
end{align*}
is a generating system of $U$:
begin{align*}
U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4
end{align*}
when $mathbb{C}^4$ is considered as $mathbb{R}$-vectorspace.
Let $(a,b,c,d)inmathbb{U}$,
Since
begin{align*}
&Re(a)(1,-1,0,0)+Im(a)(i,-i,0,0)\
&qquad+Re(a+b)(0,1,-1,0)+Im(a+b)(0,i,-i,0)\
&qquad+Re(a+b+c)(0,0,1,-1)+Im(a+b+c)(0,0,i,-i)\
&quad=(a,-a,0,0)+(0,a+b,-a-b,0)+(0,0,a+b+c,-a-b-c)\
&quad=(a,b,c,-a-b-c)\
&quad,,color{blue}{=(a,b,c,d)}
end{align*}
we conclude the set $G$ is a generating system of $U$.
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
$endgroup$
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In the same way as you already did when $mathbb{C}^4$ was considered as $mathbb{C}$-vectorspace and as indicated in the comment by @Did, we show that
begin{align*}
G={&(1,-1,0,0),(0,1,-1,0),(0,0,1,-1),\
&(i,-i,0,0),(0,i,-i,0),(0,0,i,-i)}
end{align*}
is a generating system of $U$:
begin{align*}
U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4
end{align*}
when $mathbb{C}^4$ is considered as $mathbb{R}$-vectorspace.
Let $(a,b,c,d)inmathbb{U}$,
Since
begin{align*}
&Re(a)(1,-1,0,0)+Im(a)(i,-i,0,0)\
&qquad+Re(a+b)(0,1,-1,0)+Im(a+b)(0,i,-i,0)\
&qquad+Re(a+b+c)(0,0,1,-1)+Im(a+b+c)(0,0,i,-i)\
&quad=(a,-a,0,0)+(0,a+b,-a-b,0)+(0,0,a+b+c,-a-b-c)\
&quad=(a,b,c,-a-b-c)\
&quad,,color{blue}{=(a,b,c,d)}
end{align*}
we conclude the set $G$ is a generating system of $U$.
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
$endgroup$
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
add a comment |
$begingroup$
In the same way as you already did when $mathbb{C}^4$ was considered as $mathbb{C}$-vectorspace and as indicated in the comment by @Did, we show that
begin{align*}
G={&(1,-1,0,0),(0,1,-1,0),(0,0,1,-1),\
&(i,-i,0,0),(0,i,-i,0),(0,0,i,-i)}
end{align*}
is a generating system of $U$:
begin{align*}
U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4
end{align*}
when $mathbb{C}^4$ is considered as $mathbb{R}$-vectorspace.
Let $(a,b,c,d)inmathbb{U}$,
Since
begin{align*}
&Re(a)(1,-1,0,0)+Im(a)(i,-i,0,0)\
&qquad+Re(a+b)(0,1,-1,0)+Im(a+b)(0,i,-i,0)\
&qquad+Re(a+b+c)(0,0,1,-1)+Im(a+b+c)(0,0,i,-i)\
&quad=(a,-a,0,0)+(0,a+b,-a-b,0)+(0,0,a+b+c,-a-b-c)\
&quad=(a,b,c,-a-b-c)\
&quad,,color{blue}{=(a,b,c,d)}
end{align*}
we conclude the set $G$ is a generating system of $U$.
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
$endgroup$
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
add a comment |
$begingroup$
In the same way as you already did when $mathbb{C}^4$ was considered as $mathbb{C}$-vectorspace and as indicated in the comment by @Did, we show that
begin{align*}
G={&(1,-1,0,0),(0,1,-1,0),(0,0,1,-1),\
&(i,-i,0,0),(0,i,-i,0),(0,0,i,-i)}
end{align*}
is a generating system of $U$:
begin{align*}
U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4
end{align*}
when $mathbb{C}^4$ is considered as $mathbb{R}$-vectorspace.
Let $(a,b,c,d)inmathbb{U}$,
Since
begin{align*}
&Re(a)(1,-1,0,0)+Im(a)(i,-i,0,0)\
&qquad+Re(a+b)(0,1,-1,0)+Im(a+b)(0,i,-i,0)\
&qquad+Re(a+b+c)(0,0,1,-1)+Im(a+b+c)(0,0,i,-i)\
&quad=(a,-a,0,0)+(0,a+b,-a-b,0)+(0,0,a+b+c,-a-b-c)\
&quad=(a,b,c,-a-b-c)\
&quad,,color{blue}{=(a,b,c,d)}
end{align*}
we conclude the set $G$ is a generating system of $U$.
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
$endgroup$
In the same way as you already did when $mathbb{C}^4$ was considered as $mathbb{C}$-vectorspace and as indicated in the comment by @Did, we show that
begin{align*}
G={&(1,-1,0,0),(0,1,-1,0),(0,0,1,-1),\
&(i,-i,0,0),(0,i,-i,0),(0,0,i,-i)}
end{align*}
is a generating system of $U$:
begin{align*}
U:={(a,b,c,d)|a+b+c+d=0)},Usubseteq mathbb{C}^4
end{align*}
when $mathbb{C}^4$ is considered as $mathbb{R}$-vectorspace.
Let $(a,b,c,d)inmathbb{U}$,
Since
begin{align*}
&Re(a)(1,-1,0,0)+Im(a)(i,-i,0,0)\
&qquad+Re(a+b)(0,1,-1,0)+Im(a+b)(0,i,-i,0)\
&qquad+Re(a+b+c)(0,0,1,-1)+Im(a+b+c)(0,0,i,-i)\
&quad=(a,-a,0,0)+(0,a+b,-a-b,0)+(0,0,a+b+c,-a-b-c)\
&quad=(a,b,c,-a-b-c)\
&quad,,color{blue}{=(a,b,c,d)}
end{align*}
we conclude the set $G$ is a generating system of $U$.
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
answered Jan 26 at 18:05
Markus ScheuerMarkus Scheuer
61.8k457147
61.8k457147
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
add a comment |
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
$begingroup$
@RM777: Many thanks for accepting my answer and granting the bounty.
$endgroup$
– Markus Scheuer
Jan 26 at 20:03
add a comment |
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$begingroup$
Indeed, U is an R-vector subspace of dimension 6, of the R-vector space U=C^4 of dimension 8. What is your question?
$endgroup$
– Did
Jan 19 at 19:27
$begingroup$
How can I prove that my suggested base is a generating System of $U$? So that I get every Vector of the form $(a,b,c,d)|a+b+c+d=0$ So far I have just assumed that the Dimension of U is 6 and suggested the base $ (i,-i,0,0);(0,i,-i,0);(0,0,i,-i);(1,-1,0,0);(0,1,-1,0);(0,0,1,-1)$. But I have not proved that it is indeed a generating system
$endgroup$
– RM777
Jan 19 at 19:30
$begingroup$
Well, $a+b+c+d=0$ iff $Re(a+b+c+d)=0$ and $Im(a+b+c+d)=0$. Now, $Re(a+b+c+d)=Re(a)+Re(b)+Re(c)+Re(d)$ and $Im(a+b+c+d)=Im(a)+Im(b)+Im(c)+Im(d)$ hence...
$endgroup$
– Did
Jan 19 at 19:33
1
$begingroup$
"This question has not received enough attention." Are you sure about that?
$endgroup$
– Did
Jan 20 at 21:46