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P−1 An×n P = D = ⎜ .
C 0 which implies A [P∗1 | · · · | P∗n ] = [P∗1 | 0 · · · λn
· · · | P∗ n ] ⎝ .
··· ⎞ 0
λn or, equiva- lently, [AP∗1 | · · · | AP∗n ] = [λ1 P∗1 | · · · | λn P∗n ] . Consequently, AP∗j = λj P∗j
for each j, so each (λj , P∗j ) is an eigenpair for A. In other words, P−1 AP = D
implies that P must be a matrix whose columns constitute n linearly independent eigenvectors, and D is a diagonal matrix whose diagonal entries are the
corresponding eigenvalues. It’s straightforward to reverse the above argument to
prove the converse—i.e., if there exists a linearly independent set of n eigenvectors that are used as columns to build a nonsingular matrix P, and if D is the
diagonal matrix whose diagonal entries are the corresponding eigenvalues, then
P−1 AP = D. Below is a summary. Copyright c 2000 SIAM Buy online from SIAM
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7.2 Diagonalization by Similarity Transformations
http://www.amazon.com/exec/obidos/ASIN/0898714540 507 Diagonalizability
• It is illegal to print, duplicate, or distribute this material
Please report violations to email@example.com • • A square matrix A is said to be diagonalizable whenever A is
similar to a diagonal matrix.
A complete set of eigenvectors for An×n is any set of n linearly independent eigenvectors for A. Not all matrices have complete sets of eigenvectors—e.g., consider (7.2.1) or Example 7.1.2.
Matrices that fail to possess complete sets of eigenvectors are sometimes called deﬁcient or defective matrices. D
E An×n is diagonalizable if and only if A possesses a complete set of
eigenvectors. Moreover, P−1 AP = diag (λ1 , λ2 , . . . , λn ) if and only
if the columns of P constitute a complete set of eigenvectors and
the λj ’s are the associated eigenvalues—i.e., each (λj , P∗j ) is an
eigenpair for A. T
H Example 7.2.1 IG
R Problem: If possible, diagonalize the following matrix with a similarity transform...
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