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Math Help - checking if matrix is diagonalizable

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    Member Jskid's Avatar
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    checking if matrix is diagonalizable

    My textbook gives two theorems that are used to determine if a matrix is diagonilable
    1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
    2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

    So if I have characteristic polynomial with two roots that are the same (e.g. \lambda=1 but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?
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    Quote Originally Posted by Jskid View Post
    My textbook gives two theorems that are used to determine if a matrix is diagonilable
    1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
    2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

    So if I have characteristic polynomial with two roots that are the same (e.g. \lambda=1 but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?

    Well, you have to calculate now the eigenspace of that unique eigenvalue, and the matrix is diagonalizable

    iff the eigenspace's dimension is n = the space's dimension, and non-diagonalizable otherwise.

    Tonio
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  3. #3
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    Quote Originally Posted by Jskid View Post
    My textbook gives two theorems that are used to determine if a matrix is diagonilable
    1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
    2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

    So if I have characteristic polynomial with two roots that are the same (e.g. \lambda=1 but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?
    the 2nd statement is sufficent but not required. It is just saying the if you have no repeated eigenvalues and using the fact that each eigenspace must have dimention at least 1 you will always get n linearly independent eigenvectors.

    The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.
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  4. #4
    Member Jskid's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.
    That's what happened and confused me.
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