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
)andisd
iagona
l
izab
le
,
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
,wehave
A
=
S
(
S
−
1
AS
)
S
−
1
=
SλIS
−
1
=
λSS
−
1
=
λI
1.1.2 Question.
Let
A
∈
M
n
and
f
(
t
)=
det
(
I
+
tA
)
.Wha
ti
s
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
)
²
n
±
j
=1
δ
σ
(
j
)
,j
+
t
n
°
j
=1
a
σ
(
j
)
,j
±
i
±
=
j
δ
σ
(
i
)
,i
+
o
(
t
2
)
³
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. DiFerentiating
f
(
t
)
gives
f
°
(
t
°
σ
∈
S
n
sgn
(
σ
)
²
n
°
j
=1
a
σ
(
j
)
,j
±
i
±
=
j
δ
σ
(
i
)
,i
+
o
(
t
)
³
=
n
°
j
=1
a
j,j
+
o
(
t
tr
(
A
)+
o
(
t
)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis last equality holds because
°
i
°
=
j
δ
σ
(
i
)
,i
is only nonzero when
σ
(
i
)=
i
for all
i
°
=
j
,wh
ich
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.
Amatr
ix
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
,th
enth
e
i
r
eigenspaces intersect trivially; that is
E
λ
j
∩
E
λ
k
=
{
0
}
.T
oshowth
i
s
,l
e
t
x
∈
E
λ
j
∩
E
λ
k
.Th
en
λ
j
x
=
Ax
=
λ
k
x
ichimp
l
ie
s
x
=0
because
λ
j
°
=
λ
k
h
u
si
f
{
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
2
m
j
}∪{
u
1
2
m
k
}
is a
linearly independent set, and forms a basis for
E
λ
j
+
E
λ
k
.Induc
t
ive
lyi
te
ra
t
ingth
i
s
,w
eob
ta
ina
basis for
E
λ
1
+
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '12
 BernhardBodmann
 Linear Algebra, Algebra, Matrices, Eigenvalue, eigenvector and eigenspace, diagonal matrices

Click to edit the document details