Biomedical Engineering Reference
In-Depth Information
the following result is determined:
2
4
3
5
2
4
3
5
2
4
3
5 ¼
2
4
3
5:
212
18 6 6
6 50
601
2
12
27 0 0
0 80
009
1
9
A
ð
G
Þ
¼
1
22
1
2
22
1
2
2
1
2
(A.96)
Thus, relative to the basis formed of its eigenvectors a symmetric matrix takes on
a diagonal form, the diagonal elements being its eigenvalues. This result, which was
demonstrated for a particular case, is true in general in a space of any dimension
n
as long as the matrix is symmetric.
□
There are two points in the above example that are always true if the matrix is
symmetric. The first is that the eigenvalues are always real numbers and the second
is that the eigenvectors are always mutually perpendicular. These points will now
be proved in the order stated. To prove that
is always real we shall assume that it
could be complex, and then we show that the imaginary part is zero. This proves
that
l
l
is real. If
l
is complex, say
m þ
i
n
, the associated eigenvector
t
must also
be complex and we denote it by
t
¼
p
þ
iq
. With these notations (A.48) can be
written
ð
A
fm þ
i
ng
1
Þð
p
þ
iq
Þ¼
0
:
(A.97)
Equating the real and imaginary parts, we obtain two equations,
A
p
¼ m
p
n
q
;
A
q
¼ n
p
þ m
q
:
(A.98)
The symmetry of the matrix
A
means that, for the vectors
p
and
q
,
p
A
q
q
A
p
¼
0
¼nð
p
p
þ
q
q
Þ;
(A.99)
the last equality following from taking the scalar product of the two equations in
(A.98), the first with respect to
q
and the second with respect to
p
. There is only
one-way to satisfy
n
(
p
p
þ
q
q
), since the eigenvector cannot be zero and that is
to take
is real. This result also shows that
t
must also be real.
We will now show that any two eigenvectors are orthogonal if the two associated
eigenvalues are distinct. Let
n ¼
0, hence
l
l
2
be the eigenvalues associated with the
eigenvectors
n
and
m
, respectively, then
l
1
and
A
n
¼ l
1
n
and
A
m
¼ l
2
m
:
(A.100)
Substituting the two equations (A.100) into the first and last equalities of (A.99)
we find that
ðl
1
l
2
Þ
n
m
¼
0
:
(A.101)
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