Unit VII: Symmetric Matrices
1. Spectral Theorem
A
linear transformation on a subspace
V
of
R
n
is a function
T
from
V
to
V
that satisfies
T
(
a
x
+
b
y
) =
aT
(
x
) +
bT
(
y
)
,
x
,
y
∈
V, a, b
scalars
.
The
matrix of
T
with respect to the basis
{
v
1
, . . . ,
v
m
}
of
V
is the
m
×
m
matrix
A
satisfying
T
(
v
k
) =
m
X
j
=1
A
jk
v
j
,
1
≤
k
≤
m.
These definitions are virtually the same as those given earlier for a linear transformation on
R
n
. Again note that the
k
th column of
A
contains the coordinates of
T
(
v
k
) with respect to
the basis
{
v
1
, . . . ,
v
m
}
. Also, note that if
c
1
, . . . , c
m
are the components of
u
∈
V
with respect
to the basis
{
v
1
, . . . ,
v
m
}
, and
d
1
, . . . , d
m
are the components of
T
(
u
), then
d
j
=
∑
k
A
jk
c
k
.
Lemma
If
{
u
1
, . . . ,
u
m
}
is an orthonormal basis for
V
, then the matrix
A
for
T
with
respect to the basis is given by
A
jk
=
T
(
u
k
)
·
u
j
,
1
≤
k
≤
m.
Proof.
This follows from
T
(
u
k
)
·
u
j
=
∑
r
A
rk
u
r
·
u
j
=
A
jk
.
Definition: A
symmetric linear transformation on a subspace
V
of
R
n
is a linear trans
formation
T
on
V
that satisfies
T
(
x
)
·
y
=
x
·
T
(
y
)
,
x
,
y
∈
V.
Lemma.
Let
T
be a linear transformation on a subspace
V
of
R
n
, and let
{
u
1
, . . . ,
u
m
}
be an orthonormal basis for
V
. Then
T
is a symmetric linear transformation if and only if
the matrix of
T
with respect to the basis
{
u
1
, . . . ,
u
m
}
is a symmetric matrix.
Proof.
This follows from
A
jk
=
T
(
u
k
)
·
u
j
=
u
k
·
T
(
u
j
) =
T
(
u
j
)
·
u
k
=
A
kj
.
Theorem 1.
Let
A
be a (real) symmetric
n
×
n
matrix. Then the roots of the character
istic polynomial
p
A
(
λ
) = det (
λI

A
)
of
A
are real. Thus
A
has
n
real eigenvalues (counting
multiplicity).
Proofsketch.
Let
λ
be a root of the characteristic polynomial of
A
. Then there is a nonzero
vector
z
= (
z
1
, . . . , z
n
)
∈
C
n
that satisfies (
λI

A
)
z
=
0
. We express each
z
j
=
x
j
+
iy
j
as the sum of its real and imaginary parts. Then
z
=
x
+
i
y
, where
x
= (
x
1
, . . . , x
n
) and
y
= (
y
1
, . . . , y
n
) are vectors in
R
n
. We denote
x

i
y
=
z
. Then from
z
j
z
j
=

z
j

2
, we obtain
z
·
z
=
∑

z
j

2
6
= 0. Since
A
z
=
λ
z
, we have
λ
X

z
j

2
=
X
j
λz
j
z
j
=
X
j
(
Az
)
j
z
j
=
X
j,k
A
kj
z
k
z
j
.
1
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Since the matrix
A
is symmetric and has real entries, this is equal to
X
j,k
z
k
A
jk
z
j
=
X
k
z
k
(
Az
)
k
=
X
k
z
k
λz
k
=
λ
X
k

z
k

2
.
Dividing by
∑

z
j

2
, we conclude that
λ
=
λ
, and
λ
is real.
Theorem 2.
If
T
is a symmetric linear transformation, then eigenvectors of
T
corre
sponding to different eigenvalues are orthogonal.
Proof.
If
T
(
u
) =
λ
u
and
T
(
v
) =
μ
v
, then
λ
u
·
v
=
T
(
u
)
·
v
=
u
·
T
(
v
) =
μ
u
·
v
. Thus
either
λ
=
μ
or
u
·
v
= 0.
Theorem 3.
Let
T
be a symmetric linear transformation on
R
n
. If
V
is a subspace of
R
n
such that
T
(
V
)
⊆
V
, then
T
(
V
⊥
)
⊆
V
⊥
.
Proof.
Let
w
∈
V
⊥
. If
v
∈
V
, then
T
(
v
)
∈
V
, so 0 =
T
(
v
)
·
w
=
v
·
T
(
w
). Since this
holds for all
v
∈
V
, we obtain
w
∈
V
⊥
.
Theorem 4 (Spectral Theorem).
Let
T
be a symmetric linear transformation on
R
n
.
Then there is an orthonormal basis for
R
n
consisting of eigenvectors of
T
.
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 Spring '08
 lee
 Linear Algebra, Matrices, Scalar, Orthonormal basis

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