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Math Help - Self-adjoint linear transformations and eigenvalues.

  1. #1
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    Self-adjoint linear transformations and eigenvalues.

    Hi all, I was studying for a Math test until this question stumped me:

    Let  V be an inner product space and  T : V \rightarrow V be a self-adjoint linear transformation such that  T^{2} = T .

    a) Show that all eigenvalues of  T are either 0 or 1.

    b) Describe the eigenspaces of  T in terms of the kernel of  T , the range of  T and  V .


    So for question a) I know that a self-adjoint linear transformation means  <T(u)|v> = <u|T(v)>  \forall u, v \in V and  T(u) = \lambda u where  \lambda is a scalar but I don't know how to use these to solve the question...

    As for b) I know that Nullity T + Rank T = Dim V which is equivalent to dim(ker T) + dim(im T) = dim V ...but I guess I can't solve this until I know how to do question a)...

    Any help would be greatly appreciated.
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  2. #2
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    Re: Self-adjoint linear transformations and eigenvalues.

    if T2 = T, then T satisfies the equation x2 - x = 0, which factors as x(x - 1) = 0.

    thus the minimal polynomial for T divides x2 - x, so is either:

    a) x (in which case T is the 0-matrix), and thus T has only 0 eigenvalues.

    b) x-1 (in which case T is the identity matrix), and thus T has only 1 eigenvalues,

    c) x(x-1), in which case T has both 0 and 1 for eigenvalues.

    the eigenspace corresponding to the eigenvalue 0 is called the null space (or kernel) of T. (for what is an eigenvector in this space? it is a non-zero vector v such that T(v) = 0v = 0).

    the eigenspace corresponding to the eigenvalue 1 must (in this case) be the range of T (for if for a non-zero w, we have w = T(v), then T(w) = T(T(v)) = T2(v) = T(v) = w, so w is an eigenvector with eigenvalue 1).
    Thanks from HallsofIvy
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  3. #3
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    Re: Self-adjoint linear transformations and eigenvalues.

    Quote Originally Posted by Deveno View Post
    if T2 = T, then T satisfies the equation x2 - x = 0, which factors as x(x - 1) = 0.
    So does this mean  T^{2}(x) = T(x^{2})?  Does this follow from the self-adjoint property?

    Quote Originally Posted by Deveno View Post
    thus the minimal polynomial for T divides x2 - x
    Hmm.. I'm not too sure what you mean by this. Would you mind enlightening me?

    Thank you for your explanation though, I've got a better idea of what I'm supposed to be doing now.
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  4. #4
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    Re: Self-adjoint linear transformations and eigenvalues.

    if p(x) is a polynomial, with p(x) = a0 + a1x + ....+ anxn,

    and T:V→V is a linear transformation (in particular if T is an nxn matrix that takes the nx1 matrix v to the nx1 matrix Tv)

    then if (a0I + a1T +....+ anTn)(v) = 0, for all v in V

    we say that T satisfies p(x), or that p(T) = 0.

    the monic polynomial of least degree m(x) with m(T) = 0 is called the minimal polynomial for T. it is not hard to show that if p(T) = 0, then m(x) is a factor of p(x).

    the Cayley-Hamilton theorem says that T satisfies the polynomial det(T - xI) (or, in some texts, det(xI - T), the negative of the first determinant).

    so the minimal polynomial for T, m(x), divides det(T - xI). in particular, every root of m(x) must be an eigenvalue of T.

    the converse is also true: any eigenvalue of T is also a root of the minimal polynomial.

    for suppose v is an eigenvector of T corresponding to the eigenvalue λ.

    then m(T)(v) = m(λ)v, by the same reasoning we developed in the posts above:

    (if m(x) = c0 + c1x +....+ ck-1xk-1 + xk,

    then m(T)(v) = (c0I + c1T +....+ ck-1Tk-1 + Tk)(v)

    = c0I(v) + c1T(v) +....+ ck-1Tk-1(v) + Tk(v)

    = c0v + c1(λv) +....+ ck-1k-1v) + λkv

    = (c0 + c1λ +....+ ck-1λk-1 + λk)(v) = m(λ)v, as claimed)

    but m(T) is the 0-map, by definition of the minimal polynomial, so m(λ)v = 0. since v is a non-zero vector (being an eigenvector),

    m(λ) must be 0, that is, λ is a root of m(x).

    thus the minimal polynomial tells us what all of the eigenvalues are (but perhaps not their multiplicities).

    *******

    your statement "does this mean T2(x) = T(x2)?" is meaningless.

    the "x" in the polynomial x2 - x isn't a "vector" it's an indeterminate (a "placeholder symbol" so that we can give a name to a polynomial). if you like, you can think of x as standing for a real (or complex) variable (although this is not quite accurate). we can't "square" vectors: in general, vector multiplication is undefined (we have the scalar multiplication:

    scalar times vector = vector

    and the inner product:

    vector times vector = scalar

    but we do not, in an arbitrary vector space, have a product:

    vector times vector = vector).
    Last edited by Deveno; June 9th 2012 at 07:36 AM.
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