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Math Help - Find eigenspace

  1. #1
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    Find eigenspace

    Find the eigenspace given

    A=
    0 3
    4 0

    I figured square root of 12 can the eigenvalue, not sure if 6 can also be a eigenvalue (can someone please confirm?). I then replugged the eigenvalue but got stuck row reducing the matrix to find the null space that is the eigenspace. Help please.
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  2. #2
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    how many solutions does det(A - xI) have? is 6 a solution?

    the eigenspace is not a null space, unless the eigenvalue is 0. it is the space spanned by the eigenvector(s) corresponding to a single eigenvalue.

    having found the eigenvalues, say they are λ1 and λ2, solve the equation (A-λ1I)(x,y) = 0 and then (A - λ2I)(x,y) = 0.

    these should be simple enough to solve without row-reduction.
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  3. #3
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    reduce row echelon form

    Find reduced row echelon form given matrix A=
    12^(1/2) 3
    4 12^(1/2)

    By hand I got stuck at

    1 -3/(12^(1/2))
    0 (12+48^(1/2))/(12^(1/2))

    My ti83 gives a reading of zero's in the second row, im stumped. Please help.
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  4. #4
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    Ok, so since the eigenvalue is 12^(1/2), then the eigenvector that corresponds to the eigenvector of plus minus 12^(1/2) is either:
    3/(12^(1/2)t
    t
    or
    -3/(12^(1/2)t
    t
    (I rref the matrix by ti83...cause im augmented matrix by hand challenged ><)

    Then the eigenspace is the basis of these two vectors that span given the eigenvalue.
    So if there is no eigenvalue, then there is no eigenvector and no eigenspace, but there is still either a linear independent or dependent matrix that spans or doesnt span Rn. So what did we just prove given there exist an eigenvalue/vector/space?
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  5. #5
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    the characteristic polynomial det(A - xI) is in this case: (-x)(-x)- (3)(4) = x^2 - 12.

    this has TWO solutions: x = √12 and x = -√12. forming the matrix A - √12I and setting (A - √12I)(x,y) = (0,0):

    [-√12 3][x]....[0]
    [4 -√12][y] = [0], we get the two equations:

    3y - √12x = 0
    4x - √12y = 0, both of which lead to x = √3y/2, so (√3/2, 1) is an eigenvector corresponding to the eigenvalue √12.

    thus ONE eigenspace is E√12 = span{(√3/2,1)}.

    there is another eigenspace corresponding to the eigenvalue -√12. your turn.
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  6. #6
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    Given eigenvalue = -√12

    3y + √12x = 0
    4x + √12y = 0
    x=-√3y/2
    so (-√3/2, 1) is an eigenvector corresponding to the eigenvalue -√12
    therefore the other eigenspace is E-√12 = span{(-√3/2,1)}

    This operation is a bit different from my book, because (-3/(√12),1) is equivalent to (-√3/2,1). Im using Elementary Linear Algebra by Anton & Rorres. They suggest finding the eigenspace of given eigenvalue by reduced row operation of (A-xI). Is this a misinterpretation on my part?
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  7. #7
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    reducing (A-xI) is one way to solve the system (A-xI)v = 0. when dealing with small dimensions (like 2 or 3), it is often easier to solve the system directly. for example:

    3y + √12x = 0
    4x + √12y = 0

    is equivalent to the equation:

    [√12 3][x]....[0]
    [4 √12][y] = [0]

    row reducing the matrix, multiply the first row by 4, and the second row by √12, to get:

    [4√12 12]
    [4√12 12], and now you can subtract row 1 from row 2 to get:

    [4√12 12]
    [..0... ..0.], and divide the first row by 4√12 to get:

    [1 √3/2]
    [0 ...0..], which gives the single equation x + (√3/2)y = 0, which leads to x = -√3y/2, exactly as you found.

    so the results are the same, in this case, it's just the long way around.

    and yes, when dealing with square roots, two things that look different may be the same. √3/3 looks different than 1/√3, but is in fact, the same number.
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