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Math Help - Proof of linear independence

  1. #1
    Member Mollier's Avatar
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    Proof of linear independence

    problem:

    (a)
    Prove that the four vectors
     x=(1,0,0),\;y=(0,1,0),\;z=(0,0,1)\;u=(1,1,1) in \mathbb{C}^3 form a linearly dependent set, but any three of them are linearly independent.

    (b)
     x(t)=1,\;y(t)=t,\;z(t)=t^2,\;u(t)=1+t+t^2 , prove that x,y,z,u are linearly dependent, but any three of them are linearly independent.

    attempt:

    (a) To prove that the four vectors are linearly dependent, I guess I could set up four linear equations and solve them.
    But I'm thinking it should be enough to show that:
    x+y+z-u=0.

    The thing I do not understand is the "any three of them are linearly independent" part.
    How do I go about proving that without checking all combinations?

    (b)  x+y+z-u=0. Again, I don't know how to prove the "any three of them" part without checking all combinations.

    Thanks!
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  2. #2
    Member Black's Avatar
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    For both (a) and (b), you're pretty much spot on for proving linear dependence.

    For (a), x, y, and z being independent is clear. Therefore, we need to show that u and any two of x,y, and z are linearly independent. Let x=\mathbf{e}_1, y=\mathbf{e}_2, z=\mathbf{e}_3 and let i,j \in \{1,2,3\}, where i \not= j. If we have (for a,b,c \in \mathbb{C})

    a\mathbf{e}_i+b\mathbf{e}_j+cu=0 ,

    then no matter what i and j are, we will always end up with the following system of equations:

    a+c=0

    b+c=0

    c=0,

    which implies a=b=c=0.

    Working out part (b) is very similar.


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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by Black View Post

    For (a), x, y, and z being independent is clear. Therefore, we need to show that u and any two of x,y, and z are linearly independent.

    Can I use the statement that "x,y,z being independent is clear" in a proof?

    As for (b), is this ok:

    Let x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j
    If for a,b,c \in \mathbb{C} :

     at^i+bt^j+cu=0

    then for all i and j we get a=b=c=0.

    Thanks.
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  4. #4
    Member Black's Avatar
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    Quote Originally Posted by Mollier View Post
    Can I use the statement that "x,y,z being independent is clear" in a proof?
    I guess it depends on the teacher that's grading your work. You should prove that they are independent (to be on the safe side), but it's very straightforward.

    Quote Originally Posted by Mollier View Post
    As for (b), is this ok:

    Let x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j
    If for a,b,c \in \mathbb{C} :

     at^i+bt^j+cu=0

    then for all i and j we get a=b=c=0.

    Thanks.
    Yep, pretty much. No matter what i and j are, when you group the like terms together and compare coefficients, you'll end up with the same system of equations as part (a).
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  5. #5
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    Smile

    Quote Originally Posted by Mollier View Post
    problem:

    (a)
    Prove that the four vectors
     x=(1,0,0),\;y=(0,1,0),\;z=(0,0,1)\;u=(1,1,1) in \mathbb{C}^3 form a linearly dependent set, but any three of them are linearly independent.

    (b)
     x(t)=1,\;y(t)=t,\;z(t)=t^2,\;u(t)=1+t+t^2 , prove that x,y,z,u are linearly dependent, but any three of them are linearly independent.

    attempt:

    (a) To prove that the four vectors are linearly dependent, I guess I could set up four linear equations and solve them.
    But I'm thinking it should be enough to show that:
    x+y+z-u=0.

    The thing I do not understand is the "any three of them are linearly independent" part.
    How do I go about proving that without checking all combinations?

    (b)  x+y+z-u=0. Again, I don't know how to prove the "any three of them" part without checking all combinations.

    Thanks!
    hi
    u\in Span(z,y,x)
    and (z,y,x) is a simple basis of \mathbb{C}^3 which means it's L.I.
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  6. #6
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    Smile

    Quote Originally Posted by Mollier View Post
    Can I use the statement that "x,y,z being independent is clear" in a proof?

    As for (b), is this ok:

    Let x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j
    If for a,b,c \in \mathbb{C} :

     at^i+bt^j+cu=0

    then for all i and j we get a=b=c=0.

    Thanks.
    hi
    put,
    \lambda _0+\lambda _1t+\lambda _2t^2=0
    For t=0 ,you get \lambda _0.
    Differentiate with respect to t,
    \lambda _1+2\lambda _2t=0.
    put t=0 and get \lambda _1=0.
    ....
    ...and you get \lambda _2=0.
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  7. #7
    Member Mollier's Avatar
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    Great stuff guys, thank you very much!

    Edit-
    Roah: just saw you last post, sweet!
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