Class 28
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eigenvalues and eigenvectors continued; diagonalization; multiple
eigenvalues and diagonalizability
(Sections 6.2)
Beware: we allow
eigenvalues to be complex!
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Theorem:
IF the roots of the characteristic polynomial of the matrix A are
distinct, then A is diagonalizable.
(The converse is not necessarily true!)
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Problems:
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Show that the set of k eigenvectors corresponding to k
distinct eigenvalues form a linearly independent set.
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Are the following matrices diagonalizable and if so, find the
diagonalization.
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A= [ 0 0 1 ]
[ 0 1 2 ]
[ 0 0 1 ]
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B= [ 0 0 0 ]
[ 0 1 0 ]
[ 1 0 1 ]