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=110010002
B
110
010
002
(1)
then
sI-B-1=1s-12s-2s-1s-2s-200s-1s-2000s-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-Bx=0
IB
x
0
,
i.e.,
0-10000001x1x2x3=000
0
-1
0
000
001
x1
x2
x3
0
0
0
Clearly
𝒩
λ1
I-B={c100T|c∈ℝ}
𝒩
λ1
I
B
c
100
c
Arguing along the same lines we also find
𝒩
λ2
I-B={c001T|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.
