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Math Help - How many different basis does this set contain?

  1. #1
    Member Mollier's Avatar
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    How many different basis does this set contain?

    Hi,

    problem:

    Consider the set of all those vectors in \mathbb{C}^3 each of whose coordinates is either 0 or 1. How many different basis does this set contain?

    attempt:

    The set I have to choose from has the following elements in it.

    \{(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),  (1,1,1)\}.

    I've excluded (0,0,0) as this is not part of any basis. More precisely I have
    2^3-1 elements in my set.

    Now I want to pick three vectors out of a set of seven such that they are linearly independent..Any hints on this one?

    Thanks.
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    Consider the set of all those vectors in \mathbb{C}^3 each of whose coordinates is either 0 or 1. How many different basis does this set contain?

    attempt:

    The set I have to choose from has the following elements in it.

    \{(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),  (1,1,1)\}.

    I've excluded (0,0,0) as this is not part of any basis. More precisely I have
    2^3-1 elements in my set.

    Now I want to pick three vectors out of a set of seven such that they are linearly independent..Any hints on this one?

    Thanks.
    So it seems you meant to ask: how many different basis for \mathbb{C}^3 are there in the set of all vectors all of whose coordinates are only 0 or 1...right?
    Then your set has 8 elements (the vector (0,0,0) also belongs to it), but certainly you can make it with the 7 vectors you wrote down since you want 3 lin. ind. vectors.

    Now observe that ANY two different vectors from these 7 ones are lin. independent (why?), so you need:

    (1) find out how many basis are there for any two fixed elements in your set, and then

    (2) for each set with three elements as above you have to find all the possible orderings of the vectors , since a basis in \mathbb{C}^3 is NOT only three lin. ind. vectors but in fact three ORDERED lin. ind. vectors, i.e.: the basis \{u_1,u_2,u_3\} is not the same as the basis \{u_2,u_1,u_3\} , though these two, as SETS, are identical...

    Tonio
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by tonio View Post
    So it seems you meant to ask: how many different basis for \mathbb{C}^3 are there in the set of all vectors all of whose coordinates are only 0 or 1...right?
    I actually copied the question from a book, but I think what you are writing is the same.

    Quote Originally Posted by tonio View Post
    Now observe that ANY two different vectors from these 7 ones are lin. independent (why?), so you need:
    Well, if two vectors in \mathbb{C}^3 are to be linearly dependent, they have to be colinear.
    Since we are creating different vectors with coordinates 0 or 1, two vectors in \mathbb{C}^3 must be linearly independent.
    Is this an ok answer?

    Quote Originally Posted by tonio View Post
    (1) find out how many basis are there for any two fixed elements in your set
    (2) for each set with three elements as above you have to find all the possible orderings of the vectors , since a basis in \mathbb{C}^3 is NOT only three lin. ind. vectors but in fact three ORDERED lin. ind. vectors, i.e.: the basis \{u_1,u_2,u_3\} is not the same as the basis \{u_2,u_1,u_3\} , though these two, as SETS, are identical...
    It seems that for all two fixed elements except (1,1,0),(0,1,1) (for which I get 5 basis), I get 4 basis. Funky.
    Is there a nice way of showing this?

    Also, the number of ways I can pick two vectors out of seven, no repetition, order does not matter is:
    \frac{7!}{2!(7-3)!}=21.
    So for 20 of the pairs, I have 20x4=80 basis, and for 1 of them I have 5 basis. So in total 85 basis..

    As I see it, I have 85 different basis. As you said, order does matter when it comes to basis.
    So then, each of my 85 basis has three vectors, so each basis could be rearranged in 3!=6 different ways..
    This means a whopping amount of 85x6=510 basis! I think I have done something wrong here

    Thanks tonio!
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