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Lecture

# Diagonalization

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Department
Mathematics
Course
MAT224H1
Professor
Martin, Burda
Semester
Winter

Description
Lecture 8 notes by Y. Burda 1 Diagonalization of operators, continued Last time we showed that eigenvectors corresponding to dierent eigenvalues are linearly independent. Now we will use it to get a criterion for when an operator is diagonalizable. Theorem. Let T : V V be a linear operator. Suppose that its charac- n1 nk teristic polynomial factors as p (T) = (x ) 1 ... (x k) . Then it is diagonalizable if and only if dimE = n ior every i, i.e. geometric i multiplicity coincides with algebraic multiplicity for all eigenvalues. Proof. For one direction we suppose that the operator T is diagonalizable and try to prove that dimE i = n .iSince T is diagonalizable, there is a 0 0 ... . basis v ,...,v in which its matrix is diagonal: 0 . 1 n . .. .. . . . 0 0 ... 0 Since we know the characteristic polynomial of this matrix, we get that n 1 of the entries on the diagonal are equal to ,
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