Harking back to our previous discussion of The Laplace Transform we labeled the complex
number λλ an
eigenvalue of BB
if
λI−B
λ
I
B
was not invertible. In
order to find such λλ
one has only to find those ss
for which
sI−B-1
s
I
B
is not defined. To take a concrete example we note that if
B=(
110
010
002
)
B
110
010
002
(1)
then
sI−B-1=1s−12(s−2)(
(s−1)(s−2)s−20
0(s−1)(s−2)0
00s−12
)
s
I
B
1
s
1
2
s
2
s1
s2
s2
0
0
s1
s2
0
0
0
s1
2
(2)
and so
λ1=1
λ1
1
and
λ2=2
λ2
2
are the two eigenvalues of
BB.
Now, to say that
λj
I−B
λj
I
B
is not invertible is to say that its columns
are linearly dependent, or, equivalently, that the null space
𝒩
λj
I−B
𝒩
λj
I
B
contains more than just the zero vector. We call
𝒩
λj
I−B
𝒩
λj
I
B
the
jjth
eigenspace
and call each of its nonzero members a
jjth
eigenvector. The dimension of
𝒩
λj
I−B
𝒩
λj
I
B
is referred
to as the
geometric multiplicity of
λj
λj
.
With respect to
BB above, we
compute
𝒩
λ1
I−B
𝒩
λ1
I
B
by solving
(I−B)x=0
IB
x
0
,
i.e.,
(
0-10
000
001
)(
x1
x2
x3
)=(
0
0
0
)
0
-1
0
000
001
x1
x2
x3
0
0
0
Clearly
𝒩
λ1
I−B=
c(
100
)T
c∈R
𝒩
λ1
I
B
c
100
c
Arguing along the same lines we also find
𝒩
λ2
I−B=
c(
001
)T
c∈R
𝒩
λ2
I
B
c
001
c
That
BB
is 3x3 but possesses only 2 linearly eigenvectors leads us to
speak of
BB as defective. The
cause of its defect is most likely the fact that
λ1
λ1
is a double pole of
sI−B-1
sI
B
In order to flesh out that remark and uncover the missing eigenvector
we must take a much closer look at the transfer function
Rs≡sI−B-1
R
s
sI
B
In the mathematical literature this
quantity is typically referred to as the
Resolvent of
BB.
