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Math Help - positive definite problem

  1. #1
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    positive definite problem

    3)
    A)
    there is a lenear transformation T:V->V in inner product
    it is given that T(T^{*}-3I)+2I=0. prove that T is positive definite and find every posible
    eigenvalue of T
    ?
    B)
    we define q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}
    write the formula as a sum of squares,find the rank and signature of q
    ?
    regarding A:
    the book definition is:
    a matrices which is hermitic A^{*}=A and for every vector and k of our choosing (u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0
    i dont know how to use it here
    regarding B :
    q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}
    in the books definition a signature is \sigma=\pi-\mu
    \pi is the number of positive members=2
    \mu is the number of negative members =4
    i got \sigma=-2
    ??
    in order to find the rank i need to find representative matrices in
    x'=x_{1}+x_{4}
    y'=x_{2}-3x_{3}
    and the rest stays the same
    i got here 5 member in power 2
    but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
    ??
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  2. #2
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    Re: positive definite problem

    for part B: note that: 2x_1x_4 = (\frac{x_1+x_4}{\sqrt{2}})^2 - (\frac{x_1-x_4}{\sqrt{2}})^2

    see if you can derive a similar expression for -6x_2x_3 . this should give you 2 positive terms, and 2 negative terms, for a total of 4.
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  3. #3
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    Re: positive definite problem

    Quote Originally Posted by transgalactic View Post
    3)
    A)
    there is a lenear transformation T:V->V in inner product
    it is given that T(T^{*}-3I)+2I=0. prove that T is positive definite and find every posible
    eigenvalue of T
    ?
    B)
    we define q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}
    write the formula as a sum of squares,find the rank and signature of q
    ?
    regarding A:
    the book definition is:
    a matrices which is hermitic A^{*}=A and for every vector and k of our choosing (u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0
    i dont know how to use it here
    regarding B :
    q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}
    in the books definition a signature is \sigma=\pi-\mu
    \pi is the number of positive members=2
    \mu is the number of negative members =4
    i got \sigma=-2
    ??
    in order to find the rank i need to find representative matrices in
    x'=x_{1}+x_{4}
    y'=x_{2}-3x_{3}
    and the rest stays the same
    i got here 5 member in power 2
    but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
    ??
    why this result is wrong i see that 5 members are a problem
    q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}
    but i completed the goal of transforming it into squared members only.
    whats wrong with how i have done it?
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  4. #4
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    Re: positive definite problem

    think of it like "a change of variables". the set of variables {u1 = x1+x4, u2 = -x1, u3 = -x4, u4 = x2-3x3, u5 = -x2, u6 = -9x3} is not linearly independent,

    whereas {u1 = x1+x4, u2 = x1-x4, u3 = x2+x3, u4 = x2-x3} is linearly independent.
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  5. #5
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    Re: positive definite problem

    i used a method in my book called lagrange method.
    it tells to complete the members into the (a+b)^2 formula

    i see that my variables are dependant but i coulnt think of another way of solving it

    is there a general method of solving this type of questions without getting a dependant set of variables in the end
    ?
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  6. #6
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    Re: positive definite problem

    look at post #2 again....
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  7. #7
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    Re: positive definite problem

    you have showed there the end product
    of 4 variables which are independant.

    but i have an algorithm(lagrange method) of completing to (a+b)^2
    and it brang me to a dead end solution
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  8. #8
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    Re: positive definite problem

    show the steps you took in your method, perhaps you over-looked something....
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  9. #9
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    Re: positive definite problem

    Quote Originally Posted by transgalactic View Post
    3)
    A)
    there is a lenear transformation T:V->V in inner product
    it is given that T(T^{*}-3I)+2I=0. prove that T is positive definite and find every posible
    eigenvalue of T
    ?
    B)
    we define q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}
    write the formula as a sum of squares,find the rank and signature of q
    ?
    regarding A:
    the book definition is:
    a matrices which is hermitic A^{*}=A and for every vector and k of our choosing (u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0
    i dont know how to use it here
    regarding B :
    q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}
    in the books definition a signature is \sigma=\pi-\mu
    \pi is the number of positive members=2
    \mu is the number of negative members =4
    i got \sigma=-2
    ??
    in order to find the rank i need to find representative matrices in
    x'=x_{1}+x_{4}
    y'=x_{2}-3x_{3}
    and the rest stays the same
    i got here 5 member in power 2
    but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
    ??

    i need to transform each one of these two members 2x_{1}x_{4}-6x_{2}x_{3} into to (a+b)^2 or
    (a-b)^2
    so i thought ,how do i get 2x_{1}x_{4} to be (x_{1}+x_{4})^2 .
    and i just saw that i should substract the squares to get there.
    2x_{1}x_{4}=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}
    same thing for 6x_{2}x_{3}
    6x_{2}x_{3}=(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}
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  10. #10
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    Re: positive definite problem

    is -x1^2 - x2^2 of either form (a+b)^2 or (a-b)^2? no, it is not.

    however: (a+b)^2 - (a-b)^2 = 4ab, that is:

    ab = ((a+b)/2)^2 - ((a-b)/2)^2 and these terms are of the required form (using a' = a/2, b' = b/2).

    if you have a term like:

    kab, then you use a' = a√k/2, b' = b√k/2.

    your way gves you too many variables to be able to come up with a 4x4 matrix.

    the goal is to replace each variable x1,x2,x3,x4 with some other variables u1,u2,u3,u4,

    so that q(u1,u2,u3,u4) = (c1u1)^2 (c2u2)^2 (c3u3)^3 (c4u4)^2, so that the matrix for q

    in the variables (u1,u2,u3,u4) is diag(c1,c2,c3,c4).
    Last edited by Deveno; October 3rd 2011 at 12:38 AM.
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  11. #11
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    Re: positive definite problem

    Quote Originally Posted by Deveno View Post
    is -x1^2 - x2^2 of either form (a+b)^2 or (a-b)^2? no, it is not.

    however: (a+b)^2 - (a-b)^2 = 4ab, that is:

    ab = ((a+b)/2)^2 - ((a-b)/2)^2 and these terms are of the required form (using a' = a/2, b' = b/2).

    if you have a term like:

    kab, then you use a' = a√k/2, b' = b√k/2.

    your way gves you too many variables to be able to come up with a 4x4 matrix.

    the goal is to replace each variable x1,x2,x3,x4 with some other variables u1,u2,u3,u4,

    so that q(u1,u2,u3,u4) = (c1u1)^2 (c2u2)^2 (c3u3)^3 (c4u4)^2, so that the matrix for q

    in the variables (u1,u2,u3,u4) is diag(c1,c2,c3,c4).
    so if i have 2x_1x_4-6x_2x_3
    you said that when i have kab, then i should use use a' = a√k/2, b' = b√k/2
    so for 2x_1x_4 i use x_1'=\frac{\sqrt{2}x1}{2} and
    x_4'=\frac{\sqrt{2}x4}{2}
    then i use
    by multiplying x'1 and x'4 i dont get the proper form
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  12. #12
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    Re: positive definite problem

    suppose a' = a√k/2 and b' = b√k/2.

    what is (a'+b')^2 - (a'-b')^2?
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  13. #13
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    Re: positive definite problem

    a'=\frac{{a\sqrt{k}}}{2} b'=\frac{{b\sqrt{k}}}{2}

    (a'+b')^{2}-(a'-b')^{2}=(\frac{{a\sqrt{k}}}{2}+\frac{{b\sqrt{k}}}{  2})^{2}-(\frac{{a\sqrt{k}}}{2}-\frac{{b\sqrt{k}}}{2})^{2}=2ab\sqrt{k}

    i understand
    thanks
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  14. #14
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    Re: positive definite problem

    almost....

    (a'+b')^2 - (a'-b')^2 = (\frac{a\sqrt{k}}{2}+\frac{b\sqrt{k}}{2})^2 - (\frac{a\sqrt{k}}{2}-\frac{b\sqrt{k}}{2})^2

    = (\frac{a^2k}{4} + \frac{2abk}{4} + \frac{b^2k}{4}) - (\frac{a^2k}{4} -  \frac{2abk}{4} + \frac{b^2k}{4}) = kab
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  15. #15
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    Re: positive definite problem

    regarding part A)
    in order to prove that its absolute positive we need to show that T=T^* and that all of its eigen values
    are positive

    i cant see how can i get it from T(T^{*}-3I)+2I=0
    or that its inner product

    ?
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