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574
Chapter 7
Eigenvalues and Eigenvectors
http://www.amazon.com/exec/obidos/ASIN/0898714540
7.7 NILPOTENT MATRICES AND JORDAN STRUCTURE
While it’s not always possible to diagonalize a matrix A ∈ C m×m with a similarity transformation, Schur’s theorem (p. 508) guarantees that every A ∈ C m×m
is unitarily similar to an upper-triangular matrix—say U∗ AU = T. But other
than the fact that the diagonal entries of T are the eigenvalues of A, there is
no pattern to the nonzero part of T. So to what extent can this be remedied by
giving up the unitary nature of U? In other words, is there a nonunitary P for
which P−1 AP has a simpler and more predictable pattern than that of T? We
have already made the ﬁrst step in answering this question. The core-nilpotent
decomposition (p. 397) says that for every singular matrix A of index k and
rank r, there is a nonsingular matrix Q such that
Q−1 AQ = 0
L Cr×r
0 D
E T
H , where rank (C) = r and L is nilpotent of index k. Consequently, any further simpliﬁcation by means of similarity transformations
can revolve around C and L. Let’s begin by examining the degree to which
nilpotent matrices can be reduced by similarity transformations.
In what follows, let Ln×n be a nilpotent matrix of index k so that Lk = 0
but Lk−1 = 0. The ﬁrst question is, “Can L be diagonalized by a similarity
transformation?” To answer this, notice that λ = 0 is the only eigenvalue of L
because IG
R Y
P Lx = λx =⇒ Lk x = λk x =⇒ 0 = λk x =⇒ λ = 0 (since x = 0 ). So if L is to be diagonalized by a similarity transformation, it must be the case
that P−1 LP = D = 0 (diagonal entries of D must be eigenvalues of L ), and
this forces L = 0. In other words, the only nilpotent matrix that is similar to a
diagonal matrix is the zero matrix.
Assume L = 0 from now on so that L is not diagonalizable. Since L
can always be triangularized (Schur’s theorem again), our problem boils down
to ﬁnding a nonsingular P such that P−1 LP is an u...

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