Matrix Theory, Math6304
Lecture Notes from September 6, 2012
taken by Nathaniel Hammen
Last Time (9/4/12)
Diagonalization:
conditions for diagonalization
Eigenvalue Multiplicity:
algebraic and geometric multiplicity
1
Further Review
1.1
Warmup questions
1.1.1 Question.
If
A
∈
M
n
has only one eigenvalue
λ
(of multiplicity
n
) and is diagonalizable,
what is
A
?
Answer.
Because
A
is diagonalizable, there exists an invertible
S
∈
M
n
such that
S
−
1
AS
is
diagonal. In fact, because all eigenvalues are
λ
, we have
A
=
S
(
S
−
1
AS
)
S
−
1
=
S
λ
IS
−
1
=
λ
SS
−
1
=
λ
I
1.1.2 Question.
Let
A
∈
M
n
and
f
(
t
) =
det
(
I
+
tA
)
. What is
f
(
t
)
in terms of
A
?
Answer.
The
i, j
th entry of
(
I
+
tA
)
is
δ
i,j
+
ta
i,j
, so
f
(
t
) =
det
(
I
+
tA
) =
σ
∈
S
n
sgn
(
σ
)
n
j
=1
(
δ
σ
(
j
)
,j
+
ta
σ
(
j
)
,j
)
=
σ
∈
S
n
sgn
(
σ
)
n
j
=1
δ
σ
(
j
)
,j
+
t
n
j
=1
a
σ
(
j
)
,j
i
=
j
δ
σ
(
i
)
,i
+
o
(
t
2
)
where
S
n
is the set of permutations of n elements and
sgn
(
σ
)
is +1 if
σ
is an even permutation
and 1 if
σ
is an odd permutation. Di
ff
erentiating
f
(
t
)
gives
f
(
t
) =
σ
∈
S
n
sgn
(
σ
)
n
j
=1
a
σ
(
j
)
,j
i
=
j
δ
σ
(
i
)
,i
+
o
(
t
)
=
n
j
=1
a
j,j
+
o
(
t
) =
tr
(
A
) +
o
(
t
)
1
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This last equality holds because
i
=
j
δ
σ
(
i
)
,i
is only nonzero when
σ
(
i
) =
i
for all
i
=
j
, which
only occurs when
σ
is the identity. When we plug
0
into the equation, we get that
f
(0) =
tr
(
A
)
.
1.4
Similarity (cont’d)
1.4.27 Theorem.
A matrix
A
∈
M
n
is diagonalizable if and only if the geometric and algebraic
multiplicities of each eigenvalue are equal.
Proof.
We begin by noting that if
λ
j
and
λ
k
are eigenvalues of
A
with
λ
j
=
λ
k
, then their
eigenspaces intersect trivially; that is
E
λ
j
∩
E
λ
k
=
{
0
}
. To show this, let
x
∈
E
λ
j
∩
E
λ
k
. Then
λ
j
x
=
Ax
=
λ
k
x
, which implies
x
= 0
because
λ
j
=
λ
k
. Thus if
{
v
1
, v
2
, . . . , v
m
j
}
is a basis
for
E
λ
j
and
{
u
1
, u
2
, . . . , u
m
k
}
is a basis for
E
λ
k
, then
{
v
1
, v
2
, . . . , v
m
j
}
∪
{
u
1
, u
2
, . . . , u
m
k
}
is a
linearly independent set, and forms a basis for
E
λ
j
+
E
λ
k
. Inductively iterating this, we obtain a
basis for
E
λ
1
+
· · ·
+
E
λ
r
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 Fall '12
 BernhardBodmann
 Linear Algebra, Algebra, Matrices, Eigenvalue, eigenvector and eigenspace, diagonal matrices

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