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Math Help - diagonalizability

  1. #1
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    diagonalizability

    For the following linear operator T, I have to test T for diagonalizability, and if T is diagonalizable, I have to find a basis B for V such that [T]_{B} is a diagonal matrix
    V=P_{2}(\mathbb{R}) and T is defined by T(f(x))=f(0) +f(1)(x+x^2)

    any help?
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  2. #2
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    Quote Originally Posted by dannyboycurtis View Post
    For the following linear operator T, I have to test T for diagonalizability, and if T is diagonalizable, I have to find a basis B for V such that [T]_{B} is a diagonal matrix
    V=P_{2}(\mathbb{R}) and T is defined by T(f(x))=f(0) +f(1)(x+x^2)

    any help?
    Evaluate the characteristic polynomial p_T(x) of the operator, perhaps by first finding its matrix wrt the basis \{1,x,x^2\} . You'll get p_T(x)=x(x-2)^2 , so the matrix (and thus T) are diagonalizable iff the geometric multiplicity of the eigenvalue \lambda=2 is 2.
    Check that this indeed is the case and then you're done.

    Tonio
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  3. #3
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    If you had three distinct eigenvalues, since eigenvectors corresponding to distinct eigenvalues are independent, you would know immediately that this matrix has three independent eigenvectors. Here, since there is a double eigenvalue, you will have to check if there really exist two independent eigenvector corresponding to that eigenvalue.
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  4. #4
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    sorry, how did you get the characteristic polynomial x(x-2)^2?
    I came up with (f(0)-t)(f(1)-t)^2
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Please show us what you did so we can tell you what you did wrong. Do you know the definition of the characteristic polynomial? You should evaluate \det(A-xI), where A is the matrix of the transformation with respect to your basis. There should be no such things as f(0) or f(1) in your result.
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  6. #6
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    ok so I found that the second eigenvalue, 2 has multiplicity 2, so T is indeed diagonalizable. Now I have to find eigenspaces for each eigenvalue right? Im not sure where to proceed...

    Coincidentally the matrix I got for [T]_{B} where B is the standard basis for P_{2}(\mathbb{R})
    was \left[\begin{array}{ccc}1&0&0\\0&2&0\\0&0&2\end{array}\r  ight]
    Is this correct?
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    No, that is not correct. You have

    T(1)=1+x+x^2
    T(x)=x+x^2
    T(x^2)=x+x^2
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  8. #8
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    so the matrix [T]_{B} would be:
    \left[\begin{array}{ccc}1&0&0\\1&1&1\\1&1&1\end{array}\r  ight]
    ?
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Good! Now check that you have the good characteristic polynomial. Once you have the eigenvalues, to find the eigenspace associated to the eigenvalue \lambda, solve the matrix equation (A-\lambda I)X=0 for X; this will give you a subspace of \mathbb{R}^3 which consists of the vectors whose entries are the coordinates, in the basis \beta, of the eigenvectors of T.
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  10. #10
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    ok now I am completely boggled. How did you get the char. polynomial to be x(x-2)^2?????
    I got x(x-1)(x-2)
    Am I missing something here? If the matrix A is :
    \left[\begin{array}{ccc}1&0&0\\1&1&1\\1&1&1\end{array}\r  ight]
    then A-xI is:
    \left[\begin{array}{ccc}1-x&0&0\\1&1-x&1\\1&1&1-x\end{array}\right]
    So det(A-xI) = -x(1-x)(1-x)
    How you possibly have gotten x(x-2)^2???
    Last edited by dannyboycurtis; October 29th 2009 at 11:08 PM.
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  11. #11
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    You are right. Tonio got the wrong characteristic equation and eigenvalues. (Well, it had to happen sooner or later!) They are, as you say, 0, 1, and 2. Since those are all distinct, there are three indpendent eigenvectors and, using those as basis vectors gives you a diagonal matrix.
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