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Linear Algebra - Transformations

  1. Jun 10, 2007 #1


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    1. The problem statement, all variables and given/known data
    T is a transformation from the vector space of real 2x2 matrices back to that space. T(X) = X - trans(X) (trans = transposed)
    a)Find a base for KerT and ImT
    b)Prove that T can be diagonalized.

    2. Relevant equations

    3. The attempt at a solution
    a) If X is in KerT then X = trans(X) and so KerT is the space of all 2x2 symmetric matrices and its dimension is 3. It's base is any base of that space.
    Since dimKerT = 3 , dimImT = 1 and since
    [0 1]
    [-1 0]
    is in the image, it is a base of it.

    b) obviously 0 is an eigenvalue with three linearly independant eigenvectors. It's easy to see that 2 is an eigenvalue for all the vectors in the image and so we have four linearly independent eigenvectors which means that T can be diagonalized.

    Is that right? (b) seems a little shaky to me, is there any way i can say it better?
  2. jcsd
  3. Jun 10, 2007 #2

    matt grime

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    You can say (a) better by not referring to 'it's base', which is grammatically incorrect (you got the correct possesive earlier), and misleading since it implies it has a unique base.

    (b) is fine, though you might want to mention that since 2 is a e-value, it has an e-vector, thus there is a basis of e-vectors hence the map is diagonalizable. You should also say why it is easy to see that 2 is an e-value, really, though it is clearly given by the space of antisymmetric spaces.
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