Page 119 - Numerical Methods for Chemical Engineering

108 3 Matrix eigenvalue analysis

T
A is real, symmetric (A = A)
2
As a 12 = a 21 , D = a 11 a 22 − a , and
12
,
1 2 2
λ 1,2 = (a 11 + a 22 ) ± (a 11 − a 22 ) + 4a (3.26)
2 12
2
Since (a 11 − a 22 ) + 4a 2 ≥ 0, both eigenvalues are always real in this case.
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Analytical computation of eigenvectors

[ j]
If w [ j] is an eigenvector of A for λ j , then so is cw , for any c ∈ C, since
Aw [ j] = λ j w [ j] ⇒ Acw [ j] = λ j cw [ j] (3.27)

Thus, w [ j] can have any length and still be an eigenvector for λ j . This lack of uniqueness
results from the singularity of (A − λ j I). The eigenvector w [ j] must satisfy

[ j]
(a 11 − λ j ) a 12 w 1 0

(A − λ j I)w = [ j] = = 0 (3.28)
a 21 (a 22 − λ j ) w 0
2
which yields the two equations
[ j] [ j]
(a 11 − λ j )w 1 + a 12 w 2 = 0
(3.29)
[ j] [ j]
a 21 w 1 + (a 22 − λ j )w 2 = 0
Because det(A − λ j I) = 0, these two equations are dependent, so we pick only one to
satisfy, say the first one. The second then must be satisfied automatically. As we have
[ j]
flexibility in setting the length of the eigenvector, we are free to choose |w |= 1, which
would yield the two equations
[ j] [ j]
(a 11 − λ j )w 1 + a 12 w 2 = 0
(3.30)
[ j] 2 [ j] 2

w 1 + w 2 = 1
As the second equation is nonlinear, rather than finding a unit length eigenvector, we instead
[ j]
try to find one with w = 1 that satisfies
1
[ j] [ j]
(a 11 − λ j )w + a 12 w = 0
1 2
[ j] (3.31)
w = 1
1
This linear system has a unique solution if

(a 11 − λ j ) a 12
det = (a 11 − λ j )(0) − a 12 =−a 12 = 0 (3.32)
1 0
[ j]
in which case we obtain w from the first equation,
2
(a 11 − λ j )
[ j] [ j]
(a 11 − λ j )(1) + a 12 w = 0 ⇒ w =− (3.33)
2 2
a 12
[ j] [ j]
If a 12 = 0, we instead set w = 1. Once we have computed w , we can renormalize to
2