18.03, R05
LINEAR SYSTEMS
1
Eigenvalues and eigenvectors of matrices
A
=
a
11
a
12
. . .
a
1
n
a
21
a
22
. . .
a
2
n
.
.
.
.
.
.
.
.
.
a
n
1
a
n
2
. . .
a
nn
1. The trace of
A
is the sum of the elements on the diagonal, Tr
A
=
a
11
+
a
22
+
. . .
+
a
nn
.
2. The determinant of
A
is computed by expanding along a row or a column and keep doing
it until we reduce the computation to 2
×
2 determinants. For instance, expanding along
the
i
th row we get
det
A
=
a
i
1
(

1)
i
+1
det
m
i
1
+
a
i
2
(

1)
i
+2
det
m
i
2
+
. . .
+
a
in
(

1)
i
+
n
det
m
in
,
where
m
ij
stands for the (
i, j
) minor of
A
, namely the (
n

1)
×
(
n

1) matrix obtained
from
A
by erasing the
i
th row and the
j
th column.
3.
A
is invertible if and only if det
A
6
= 0. In that case
A

1
=
1
det
A
m
11

m
12
. . .
(

1)
1+
n
m
1
n

m
21
m
22
. . .
(

1)
2+
n
m
2
n
.
.
.
.
.
.
.
.
.
(

1)
n
+1
m
n
1
(

1)
n
+2
m
n
2
. . .
m
nn
T
,
where [ ]
T
stands for the
transpose
of a matrix and the
m
ij
’s denote minors like above.
4. The characteristic polynomial of
A
is the degree
n
polynomial
p
A
(
λ
) = det(
A

λI
n
)
.
5. The eigenvalues of
A
are the roots of
p
A
(
λ
) = 0. Counting multiplicities there are
n
of
them.
6. An eigenvector corresponding to the eigenvalue
λ
of the matrix
A
is a
nonzero
vector
v
such that
A
v
=
λ
v
, or equivalently, (
A

λI
n
)
v
=
0
.
7. Assume
λ
is a repeated eigenvalue of
A
with multiplicity
m
. It is called
complete
if there
are
m
linearly independent eigenvectors corresponding to it and
defective
otherwise.
8. If
λ
is a defective eigenvalue and
v
is a corresponding eigenvector, a
generalized eigenvector
is a vector
u
such that (
A

λI
)
u
=
v
.
1
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View Full Document18.03, R05
2
Matrix exponentials
Deﬁnition
The exponential of a square matrix
A
is
e
A
=
∞
X
k
=0
1
k
!
A
k
.
Note:
•
If
B
in another
n
×
n
matrix, then
e
A
+
B
=
e
A
e
B
if and only if
AB
=
BA
.
•
If
A
has eigenvalues
λ
1
, . . . , λ
n
, the eigenvalues of its exponential
e
A
are
e
λ
1
, . . . , e
λ
n
.
How to compute
•
Probably the easiest way is to make a system
x
0
=
A
x
and ﬁnd one of fundamental matrices
F
(
t
). Then
e
A
=
F
(1)
F
(0)

1
.
•
There is one other trick that might help with computations, namely if all the elements on
the main diagonal are equal to
r
, write
A
=
rI
+
B
. Then
B
to some power gives the zero
matrix and its exponential is easy to compute being a ﬁnite sum and
e
A
=
e
r
e
B
.
•
D
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 Spring '11
 Prof.Ramachandran
 Linear Algebra, Matrices, general solution

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