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Math Help - Eigenvectors/value proofs

  1. #1
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    Eigenvectors/value proofs

    I have a few eigenvector/value proofs and not really sure how to go about/start them, if someone could give me some ideas that would be great sorry if the notation in this post is bad, if you got any question about it i'll try to explain it a little better

    show that if A is a nxn Hermitian matrix then all its eigenvalues are real

    Show that any nonzero column vector v, v*v is a positive number

    Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m, that correspond to distinct eigenvalues is independent
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    Quote Originally Posted by action259 View Post

    Show that any nonzero column vector v, v*v is a positive number
    If you have,
    v=[a1 a2 a3 .... an]
    Given that it is nonzero, i.e. a1,a2,... not equal to zero.
    Then,
    v*v=(a1)^2+(a2)^2+...+(an)^2 sum of square is positive.
    All of them cannot be zero.
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  3. #3
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    Quote Originally Posted by action259 View Post
    I have a few eigenvector/value proofs and not really sure how to go about/start them, if someone could give me some ideas that would be great sorry if the notation in this post is bad, if you got any question about it i'll try to explain it a little better

    show that if A is a nxn Hermitian matrix then all its eigenvalues are real
    Let A be hermitian, and s be an eigen-value corresponding to eigen vector X.

    Then:

    AX=sX

    So:

    X*AX=sX*X,

    but X*X=sum(conj(x_i).x_i, i=1,..n) is real, so:

    X*AX/(X*X)=s

    and:

    conj(X*AX/(X*X))=conj(X*AX)/(X*X)=(X*A*X)/(X*X)=conj(s).

    But A is hermitian so A*=A

    so:

    conj(s)=(X*AX)/(X*X)=s,

    hence s is real.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by action259 View Post
    Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m, that correspond to distinct eigenvalues is independent
    Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m,
    that correspond to distinct eigenvalues is independent

    Suppose m=2, and that X1, and X2 are the e-vectors corresponding to
    e-values lambda1, lambda2 and that lambda1 != lambda2, further suppose
    that X1 and X2 are not linearly independent. Then there exist constants
    a, b != 0 such that:

    a.X1 + b.X2 = 0 ....(1)

    Now:

    A(a.X1+b.X2)=a.lambda1.X1 + b.lambda2.X2=0

    from (1) we have a.X1=-b.X2, so:

    (lambda1-lambda2).b.X2=0,

    but as b != 0, X2 !=0 and lambda1 != lambda2 this is
    a contradiction, so we have the desired result is true
    when m=2.

    The above constitutes the base case for an inductive
    proof that the result is true for any m<=n.

    Suppose X(i), i=1..k, k<n-1 are linearly independent e-vectors
    corresponding to distinct e-values lambda(i), i=1,..k. Further
    suppose that X(i), i=1..k+1, k<n-1 are not linearly independent
    e-vectors corresponding to distinct e-values lambda(i), i=1,..k+1.

    Then there exist constants a(i), i=1,..k+1 not all zero such that:

    sum(i=1,k+1) a(i).X(i) = 0,

    so:

    a(k+1).X(k+1) = - sum(i=1,k) a(i).X(i)...(2)

    Now multiply this by A:

    lambda(k+1).a(k+1).X(k+1) = - sum(i=1,k) a(i).lambda(i).X(i).

    Substitute from (2):

    lambda(k+1).sum(i=1,k) a(i).lambda(i).X(i)= - sum(i=1,k) a(i).lambda(i).X(i)

    or:

    sum(i=1,k) a(i).(lambda(i)-lambda(k+1)).X(i)=0.

    but as lambda(k+1) != lambda(i), i=1,..k this implies that the X(i) i=1,..k
    are not linearly independent - a contradiction.

    This last result together with the base case proves by the desired
    result by induction (it does not prove it true for all m in this case
    as we have at most n distinct e-values)

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Show that any set of m eigenvectors of an nxn matrix A, n> (or equal) m,
    that correspond to distinct eigenvalues is independent

    Suppose m=2, and that X1, and X2 are the e-vectors corresponding to
    e-values lambda1, lambda2 and that lambda1 != lambda2, further suppose
    that X1 and X2 are not linearly independent. Then there exist constants
    a, b != 0 such that:

    a.X1 + b.X2 = 0 ....(1)

    Now:

    A(a.X1+b.X2)=a.lambda1.X1 + b.lambda2.X2=0

    from (1) we have a.X1=-b.X2, so:

    (lambda1-lambda2).b.X2=0,

    but as b != 0, X2 !=0 and lambda1 != lambda2 this is
    a contradiction, so we have the desired result is true
    when m=2.

    The above constitutes the base case for an inductive
    proof that the result is true for any m<=n.

    Suppose X(i), i=1..k, k<n-1 are linearly independent e-vectors
    corresponding to distinct e-values lambda(i), i=1,..k. Further
    suppose that X(i), i=1..k+1, k<n-1 are not linearly independent
    e-vectors corresponding to distinct e-values lambda(i), i=1,..k+1.

    Then there exist constants a(i), i=1,..k+1 not all zero such that:

    sum(i=1,k+1) a(i).X(i) = 0,

    so:

    a(k+1).X(k+1) = - sum(i=1,k) a(i).X(i)...(2)

    Now multiply this by A:

    lambda(k+1).a(k+1).X(k+1) = - sum(i=1,k) a(i).lambda(i).X(i).

    Substitute from (2):

    lambda(k+1).sum(i=1,k) a(i).lambda(i).X(i)= - sum(i=1,k) a(i).lambda(i).X(i)

    or:

    sum(i=1,k) a(i).(lambda(i)-lambda(k+1)).X(i)=0.

    but as lambda(k+1) != lambda(i), i=1,..k this implies that the X(i) i=1,..k
    are not linearly independent - a contradiction.

    This last result together with the base case proves by the desired
    result by induction (it does not prove it true for all m in this case
    as we have at most n distinct e-values)

    RonL
    And think there is an easier way.
    Okay, I never tood a course in Linear Algebra (especially when it is already advanced) but I would like to make a recommendation that every finite dimensional vector space has a finite basis. Thus, maybe you use that here. Since the superset here is a basis (finite) then its subset surly contains a finite span set. Q.E.D.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    And think there is an easier way.
    Okay, I never tood a course in Linear Algebra (especially when it is already advanced) but I would like to make a recommendation that every finite dimensional vector space has a finite basis. Thus, maybe you use that here. Since the superset here is a basis (finite) then its subset surly contains a finite span set. Q.E.D.
    I'm pretty sure that there is a proof that on some measure might be
    considered simpler, but I don't see that what you are saying contributes to
    that. The idea in the proof given is in fact pretty simple if a bit fussy to
    prove.

    That these e-vectors span a subspace is not in question, what is is what
    is its dimension?


    Also Adding QED to the end of a half formed idea does not make a proof.

    RonL
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