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    Diagonalization of Linear Operator

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    Consider the linear operator T:R^3 given by T (see attached)

    Determine the eigenvectors and the corresponding eigenvalues of T. If T diagonalizable? Why or why not?

    © BrainMass Inc. brainmass.com December 24, 2021, 11:35 pm ad1c9bdddf
    https://brainmass.com/math/linear-algebra/diagonalization-linear-operator-583112

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Solution:
    Since , the matrix of this linear operator is .
    The eigenvalues we find from the equation :

    There are two different eigenvalues: and .
    Let's find the corresponding eigenvectors.
    For the eigenvector have to be solution of the homogeneous system of equations with matrix . Using Gauss method we bring this matrix to the form

    The solution of the system is .
    So there is only one linearly independent eigenvector .
    For the eigenvector have to be solution of the homogeneous system of equations with matrix . Using Gauss method we bring this matrix to the form

    The solution of the system is .
    So, again, there is only one linearly independent eigenvector .
    Totally, we have found only two linearly independent eigenvectors, which means that the linear operator is not diagonalizable.
    We may explain it also by the following argument: the algebraic multiplicity of eigenvalue is (the power of in the polynomial ), but its geometric multiplicity is (the number of linearly independent eigenvectors for this ). Since there is an eigenvalue for which algebraic and geometric multiplicities are not equal, the linear operator is not diagonalizable.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:35 pm ad1c9bdddf>
    https://brainmass.com/math/linear-algebra/diagonalization-linear-operator-583112

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