Biomedical Engineering Reference
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These three conditions, only two of which are independent, gave t 1 ¼
t 3 and
t 1 ¼
2 t 2 , leaving an undetermined parameter in the eigen n-tuple t . Now that t is a
vector, we can specify the length of a vector. Another consequence of the symmetry
of A is that these eigenvectors are orthogonal if the eigenvalues are distinct. Hence,
if we set the length of the eigenvectors to be one to remove the undetermined
parameter, we will generate an orthonormal basis from the set of three eigenvectors,
since the eigenvalues are distinct. If we use the normality condition t 1 þ
t 2 þ
t 3 ¼
1
and the results that follow from (A.56), t 1 ¼
t 3 and t 1 ¼
2 t 2 , one finds that
ð
1
3
t
¼
2 e 1 þ
e 2 þ
2 e 3 Þ
(A.91)
which shows that both t and
t are eigenvectors. This will be true for any
eigenvector because they are really eigen-directions. For the second and third
eigenvalues, 18 and 9, we find that
ð
ð
1
3
1
3
t
¼
e 1
2 e 2 þ
2 e 3 Þ
and t
¼
2 e 1
2 e 2
e 3 Þ;
(A.92)
respectively. It is easy to see that these three eigenvectors are mutually orthogonal.
It was noted above that, since the eigenvectors constitute a set of three mutually
perpendicular unit vectors in a three-dimensional space, they might be used to form
a basis or a coordinate reference frame. Let the three orthogonal eigenvectors be the
base vectors e I , e II , and e III of a Greek reference frame. From (A.91) and (A.92) we
form a new reference basis for the example eigenvalue problem, thus
ð
ð
1
3
1
3
e I ¼
2 e 1 þ
e 2 þ
2 e 3 Þ;
e II ¼
e 1
2 e 2 þ
2 e 3 Þ;
ð
1
3
e III ¼
2 e 1
2 e 2
e 3 Þ:
(A.93)
It is easy to verify that both the Greek and Latin base vectors form right-handed
orthonormal systems. The orthogonal matrix Q for transformation from the Latin to
the Greek system is given by (A.64) and (A.93) as
2
3
2
12
1
3
4
5 :
Q
¼½
Q i a ¼
1
2
2
(A.94)
22
1
Substituting the Q of (A.94) and the A specified by (A.56) into the second of
(A.83) with T
¼
A ,
A ð G Þ ¼
Q T
A ð L Þ
Q
;
(A.95)
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