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Math Help - T saved subspaces question

  1. #1
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    T saved subspaces question

    T:R^{3}->R^{3}
    A=\left(\begin{array}{ccc}2 & 1 & 0\\0 & 2 & 0\\0 & 0 & 3\end{array}\right)
    A represents T in starndart basis
    find 4 subspaces which are “T saved” (dont knpw the proper term
    T saved means if v inside W subspace then Tv in W too
    i have found two eigenvectors one for eigen value 2
    and another eigen vector for eigen value 3.
    span of each one of those eigen vectors makes Tsaved subspace.
    also this A is not diagoisable.
    so how do i find the other two T saved subspaces?
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  2. #2
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    Re: T saved subspaces question

    clearly the subspace generated by the basis {(1,0,0)} is T-invariant, since that is E2.
    also the subspace generated by {(0,0,1)} is T-invariant, since that is E3.
    therefore the subspace generated by the basis {(1,0,0),(0,0,1)} is also T-invariant. a vector in this subspace is of the form (a,0,b),
    and T(a,0,b) = (2a,0,3b) = 2a(1,0,0) + 3b(0,0,1).
    two more obvious T-invariant subspaces are R^3 itself, and the 0-subspace.
    a less obvious T-invariant subspace is the subspace generated by the basis {(1,0,0),(0,1,0)}. any vector in this subspace is of the form (a,b,0),
    and T(a,b,0) = (2a+b,2b,0) = (2a+b)(1,0,0) + 2b(0,1,0).
    so that's 6, you should be covered.
    Last edited by Deveno; October 15th 2011 at 05:17 PM.
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  3. #3
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    Re: T saved subspaces question

    ok i understand more

    regarding
    "
    a less obvious T-invariant subspace is the subspace generated by the basis {(1,0,0),(0,1,0)}. any vector in this subspace is of the form (a,b,0),

    and T(a,b,0) = (2a+b,2b,0) = (2a+b)(1,0,0) + 2b(0,1,0).
    "

    how you thought about sp{e2,e3}
    ?

    a vector from this span is a(0,1,0)+b(1,0,0)=(b,a,0)
    why its variant??

    why sp{e2,e1} will not work
    ?
    Last edited by transgalactic; October 15th 2011 at 05:20 PM.
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  4. #4
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    Re: T saved subspaces question

    span({(1,0,0),(0,1,0)}) IS span({e1,e2}), which is the same as span({e2,e1}).

    span({e2,e3}) will not work. every vector in span({e2,e3}) has 0 as a first coordinate and is of the form (0,y,z) = y(0,1,0) + z(0,0,1) = ye2 + ze3.

    but T(0,y,z) = (y,2y,3z) which is NOT in span({e2,e3}) if y is non-zero.

    the more interesting question is: how did i know span({e1,e2}) would work?

    answer: decompose T into Jordan blocks. you have a 2x2 and a 1x1. the 1x1 block corresponds to the eigenspace for the eigenvalue 3.

    the 2x2 block corresponds to span({e1,e2)}, which contains the eigenspace for the eigenvalue 2 (since T is not diagonalizable,

    this eigenvalue is "deficient", it has algebraic multiplicity 2, but geometric multiplicity 1).

    note that e2 is not in ker(T-2I), but it IS in ker((T-2I)^2), it is a "generalized eigenvector".
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  5. #5
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    Re: T saved subspaces question

    ok so
    for k=2 we got e1 so sp{e1} is invariant
    for k=3 we got e3 so sp{e3} is invariant
    so sp{e1,e3} is invariant too.

    but regarding the last one
    we know the the geometric multiplicity is less then the algebric

    so in order to find the missing one you calculated ker((T-2I)^2)
    did you actually calculated (A-2I)(A-2I) ?
    why power of 2?
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  6. #6
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    Re: T saved subspaces question

    because the size of the Jordan block is 2x2. the "block submatrix"

    [0 1]
    [0 0]

    (the block you get when you take the Jordan block corresponding to the eigenvalue 2, and subtract 2I) has nilpotency 2.

    look at it this way, we know that we can decompose R^3 into span({e1,e2})⊕span({e3)}):

    in this direct-sum decomposition, the matrix for T looks like:

    [J 0]
    [0 3], where J is the matrix

    [2 1]
    [0 2]. writing an element of R^3 as V + z, where V is in span({e1,e2}) and z is in span({e3)} (sort of an abuse of notation, we should really write (0,0,z)),

    we get T(x,y,z) =

    [J 0][V]....[JV]
    [0 3][z] = [3z]

    here, explicitly JV is the vector (2x+y,2y,0) = (2x+y)e1+2ye2. do you see how the Jordan blocks neatly "slice up" the vector space?

    but if it makes you feel better: T-2I =

    \begin{bmatrix}0&1&0\\0&0&0\\0&0&1\end{bmatrix} and (T-2I)(x,y,z) = (y,0,z)

    so ker(T-2I) is all vectors of the form (x,0,0) = span({e1}), while (T-2I)^2 =

    \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}, so (T-2I)^2(x,y,z) = (0,0,z),

    so ker(T-2I)^2 is all vectors of the form (x,y,0) = span({e1,e2}).

    in general, often you find a "repeated root" of the characteristic polynomial, and the eigenspace doesn't have enough eigenvectors

    to make a basis for V, but you still want to make the T-invariant subspaces "as big as possible". so you want to find "blocks"

    on which T is invariant. Jordan blocks give you a way to do that.
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  7. #7
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    Re: T saved subspaces question

    ok so e1 comes from ker(T-2I)
    e2 comes from ker(T-2I)^2

    how do you prove that sp{e1,e2} is variant
    ?

    if i take v form this sp{e1,e2} then in the
    {e1,e2,e3} basis it will look as
    v=av1+av2
    so
    Tv=aTv1+aTv2
    v1 is eigen vector T but v2 doesnt so we cant show that Tv belongs to sp{e1,e2} either

    ??
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  8. #8
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    Re: T saved subspaces question

    you can explicitly calculate T(ae1 + a2e2) and show that the third coordinate is 0.

    look, if v is an eigenvector for T, then span({v}) will be a T-invariant subspace.

    that doesn't mean every T-invariant subspace is determined by an eigenvector.

    don't confuse "if" conditions with "if and only if".
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