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Math Help - triangularize the operator

  1. #1
    jax
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    triangularize the operator

    +riangularize the operator T: C^3 --> C^3 given by T(x,y,z)= (x, yi, y+zi). Find the ordered basis B and the upper triangular matrix Mat(T, B ,B).

    This is what I have so far, please let me know how it looks. Thanks!


    Standard basis : (1,0,0)=(1,0,0)
    (0,1,0)=(0, i, 1)
    (0,0,1)=(0,0, i)
    Apply i times the second row and add to the third.
    Now I have (1,0,0)
    (0, i, 0)
    (0,1, i)
    So the eigen values are {1,i}
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  2. #2
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    Quote Originally Posted by jax View Post
    +riangularize the operator T: C^3 --> C^3 given by T(x,y,z)= (x, yi, y+zi). Find the ordered basis B and the upper triangular matrix Mat(T, B ,B).

    This is what I have so far, please let me know how it looks. Thanks!


    Standard basis : (1,0,0)=(1,0,0)
    (0,1,0)=(0, i, 1)
    (0,0,1)=(0,0, i)


    What do you mean with these weird equalities?? The 2nd and 3rd. ones are obviously false...

    Take simply the standard basis \left\{(1,0,0), (0,1,0), (0,0,1)\right\}...


    Apply i times the second row and add to the third.
    Now I have (1,0,0)
    (0, i, 0)
    (0,1, i)
    So the eigen values are {1,i}

    The matrix representing T wrt the standard basis above is just \begin{pmatrix}1&0&0\\0&i&0\\0&1&i\end{pmatrix} , and from it is immediate that the eigenvalues

    of the matrix are, indeed, 1,i ,as you said. Now, the matrix is diagonalizable iff the eigenspace belonging to i has dimension 2...

    Tonio
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