Mixing
Count the number of bases of the vector space $mathbb{C}^3$.
$begingroup$
Problem: Consider the set of all those vectors in $mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?
In general, if $B$ is the set of all bases vectors then,
$$B={(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)}.$$
There are $8(6cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated.
linear-algebra combinatorics
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
$endgroup$
add a comment |
$begingroup$
Problem: Consider the set of all those vectors in $mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?
In general, if $B$ is the set of all bases vectors then,
$$B={(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)}.$$
There are $8(6cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated.
linear-algebra combinatorics
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
$endgroup$
add a comment |
$begingroup$
Problem: Consider the set of all those vectors in $mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?
In general, if $B$ is the set of all bases vectors then,
$$B={(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)}.$$
There are $8(6cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated.
linear-algebra combinatorics
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
$endgroup$
Problem: Consider the set of all those vectors in $mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?
In general, if $B$ is the set of all bases vectors then,
$$B={(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)}.$$
There are $8(6cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated.
linear-algebra combinatorics
linear-algebra combinatorics
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
asked Jun 9 '17 at 13:38
Hello_WorldHello_World
4,14121931
4,14121931
add a comment |
add a comment |
2 Answers
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Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each. Since they must be different you only have $binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
$endgroup$
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
add a comment |
$begingroup$
The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \
(0, 0, 1), (1, 0, 0), (1, 0, 1) \
(0, 0, 1), (1, 1, 0), (1, 1, 1) \
(0, 1, 0), (1, 0, 0), (1, 1, 0) \
(0, 1, 0), (1, 0, 1), (1, 1, 1) \
(0, 1, 1), (1, 0, 0), (1, 1, 1) \
end{eqnarray}
$$
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each. Since they must be different you only have $binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
$endgroup$
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
add a comment |
$begingroup$
Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each. Since they must be different you only have $binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
$endgroup$
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
add a comment |
$begingroup$
Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each. Since they must be different you only have $binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
$endgroup$
Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each. Since they must be different you only have $binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
answered Jun 11 '17 at 4:47
deinstdeinst
4,4262029
4,4262029
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
add a comment |
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
$begingroup$
Additionally, it's not hard to show that three distinct non-zero $(0,1)$-vectors are linearly dependent iff they form a solution to $x+y=z$.
$endgroup$
– Erick Wong
Jun 11 '17 at 4:57
add a comment |
$begingroup$
The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \
(0, 0, 1), (1, 0, 0), (1, 0, 1) \
(0, 0, 1), (1, 1, 0), (1, 1, 1) \
(0, 1, 0), (1, 0, 0), (1, 1, 0) \
(0, 1, 0), (1, 0, 1), (1, 1, 1) \
(0, 1, 1), (1, 0, 0), (1, 1, 1) \
end{eqnarray}
$$
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
add a comment |
$begingroup$
The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \
(0, 0, 1), (1, 0, 0), (1, 0, 1) \
(0, 0, 1), (1, 1, 0), (1, 1, 1) \
(0, 1, 0), (1, 0, 0), (1, 1, 0) \
(0, 1, 0), (1, 0, 1), (1, 1, 1) \
(0, 1, 1), (1, 0, 0), (1, 1, 1) \
end{eqnarray}
$$
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
add a comment |
$begingroup$
The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \
(0, 0, 1), (1, 0, 0), (1, 0, 1) \
(0, 0, 1), (1, 1, 0), (1, 1, 1) \
(0, 1, 0), (1, 0, 0), (1, 1, 0) \
(0, 1, 0), (1, 0, 1), (1, 1, 1) \
(0, 1, 1), (1, 0, 0), (1, 1, 1) \
end{eqnarray}
$$
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
$endgroup$
The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \
(0, 0, 1), (1, 0, 0), (1, 0, 1) \
(0, 0, 1), (1, 1, 0), (1, 1, 1) \
(0, 1, 0), (1, 0, 0), (1, 1, 0) \
(0, 1, 0), (1, 0, 1), (1, 1, 1) \
(0, 1, 1), (1, 0, 0), (1, 1, 1) \
end{eqnarray}
$$
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
answered Jan 20 at 21:17
Max HerrmannMax Herrmann
704418
704418
add a comment |
add a comment |
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