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Eigenvector matrix decomposition and basis sets 117

Eigenvector matrix decomposition and basis sets

Foragiven N × N matrix, when does the set of eigenvectors form a complete basis such
N
that any v ∈ C can be expressed as a linear combination of eigenvectors?
Can we express a matrix as a decomposition involving matrices that comprise eigenvalues
and eigenvectors?
Are there special classes of matrices for which the eigenvalues can be proved to be real, or
for which the set of eigenvectors is not only linearly independent, but also orthogonal?
Such questions may seem abstract, but they are in fact very important in practice. Above,
our proof of convergence of Jacobi’s method is based upon the assumed existence of a
complete eigenvector basis for B −1 (B − A).

Eigenvector properties of a general N × N complex matrix
Because we cannot assume that the eigenvectors and eigenvalues are real, even for a real
matrix, we consider the general case where λ k ∈ C, w [k] ∈ C [N] , and A is an N × N complex
matrix,

(a 11 + ib 11 ) (a 12 + ib 12 ) ... (a 1N + ib 1N )
 
(a 21 + ib 21 ) (a 22 + ib 22 ) ... (a 2N + ib 2N )
 
  (3.85)
. . . 
. . .
A = 
 . . . 
(a N1 + ib N1 )(a N2 + ib N2 ) ... (a NN + ib NN )
If we do not make any special assumptions about the structure of the matrix, we cannot
N
establish generally that the eigenvectors form a complete basis for C . We can prove the
following statements, however.
[k]
Theorem ENVG1 Eigenvectors w [ j] and w , satisfying Aw [ j] = λ j w [ j] and Aw [k] =
λ k w [k] respectively, are linearly independent if λ j = λ k . Thus, the eigenvectors of any N ×
N
N matrix A that has N distinct eigenvalues form a complete basis for C .
[k]
Proof We want to show that when λ j = λ k , we cannot have cw [ j] = w . Let us assume,
[k]
contrary to the theorem, that cw [ j] = w . We replace w [k] with cw [ j] in the first term of
Aw [k] − λ k w [k] = 0,
Acw [ j] − λ k w [k] = λ j cw [ j] − λ k w [k] = 0 (3.86)

and replace w [k] with cw [ j] in the second term,

λ j cw [ j] − λ k cw [ j] = c(λ j − λ k )w [ j] = 0 (3.87)

This poses a contradiction with w [ j] = 0 when λ j = λ k and thus we cannot have cw [ j] =
[k]
w . As w [ j] and w [k] cannot be parallel, they are linearly independent. Thus, if A has N
[1]
distinct eigenvalues, no two of {w ,..., w [N] } can be parallel and the eigenvectors form
a complete basis set. QED