Principles of Mathematics in Operations Research

56 4 Eigen Values and Vectors

Example 4.3.3 From the previous example,

A =

MO

i 1 0

0 I 3

"44

=*Ai = 1, A 3 =

Ax = X\x <&

\x\ + |x2 = 0 <£> x\ + x 2 = 0

I x 2 + |x 3 = 0 & x 2 + x 3 = 0

±X1

\x\ + x 2

;X 2 + fx 3 _

=

1*1"

1*2

1*3.

:}

Ax = X 2 x <=>

|*i

5X1 + X 2

fX 2 |*3

Thus, v\ =

Xi

x 2

X3

7*2 - 7X3 = 0 <4> X 2

Ax = A 3 x <=>

H = 0.1

• X 3 = 0. J
  Thus, v 2 =

1*1

\x\ + x 2

[X2 + fx 3 _

=

"fail"

fx 2

.1*3.

xx =0.

O- §xi - \x 2 = 0 =>• 2a:i + x 2 = 0. \ Thus, v 3 =

x 2 = 0.

Therefore, S =

[S\I]

Then,

100

-1 10

1 1 1

100

-1 10

111

100'

010

001

-4

"100

010

0 1 1

100'

1 10

-10 1

-»

'100

010

001

1 00"

1 10

-2-1 1

= [I\S -ii

S^AS ••

1 00'

1 10

2 -1 1

1 00"

\ 10

0 1 2

."44.

100'

-1 10

111

=

loo"

01 0

oof

= A.

Remark 4.3.4 Any matrix with distinct eigen values can be diagonalized.
However, the diagonalization matrix S is not unique; hence neither is the
basis {«}"_!• If we multiply an eigen vector with a scalar, it will still remain
an eigen vector. Not all matrices posses n linearly independent eigen vectors;
therefore, some matrices are not dioganalizable.