Symmetric matrices and the second derivative test
1
Chapter 4
Symmetric matrices and the second
derivative test
In this chapter we are going to finish our description of the nature of nondegenerate critical
points. But first we need to discuss some fascinating and important features of square matrices.
A.
Eigenvalues and eigenvectors
Suppose that
A
= (
a
ij
) is a fixed
n
×
n
matrix. We are going to discuss linear equations
of the form
Ax
=
λx,
where
x
∈
R
n
and
λ
∈
R
. (We sometimes will allow
x
∈
C
n
and
λ
∈
C
.) Of course,
x
= 0
is always a solution of this equation, but not an interesting one.
We say
x
is a
nontrivial
solution if it satisfies the equation and
x
6
= 0.
DEFINITION.
If
Ax
=
λx
and
x
6
= 0, we say that
λ
is an
eigenvalue
of
A
and that the
vector
x
is an
eigenvector
of
A
corresponding to
λ
.
EXAMPLE.
Let
A
=
0
3
1
2
¶
. Then we notice that
A
1
1
¶
=
3
3
¶
= 3
1
1
¶
,
so
1
1
¶
is an eigenvector corresponding to the eigenvalue 3. Also,
A
3

1
¶
=

3
1
¶
=

3

1
¶
,
so
3

1
¶
is an eigenvector corresponding to the eigenvalue

1.
EXAMPLE.
Let
A
=
2
1
0
0
¶
. Then
A
1
0
¶
= 2
1
0
¶
, so 2 is an eigenvalue, and
A
1

2
¶
=
0
0
¶
, so 0 is also an eigenvalue.
REMARK.
The German word for eigenvalue is
eigenwert
. A literal translation into English
would be “characteristic value,” and this phrase appears in a few texts. The English word
“eigenvalue” is clearly a sort of half translation, half transliteration, but this hybrid has stuck.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
Chapter 4
PROBLEM 4–1.
Show that
A
is invertible
⇐⇒
0 is not an eigenvalue of
A
.
The equation
Ax
=
λx
can be rewritten as
Ax
=
λIx
, and then as (
A

λI
)
x
= 0. In order
that this equation have a
nonzero
x
as a solution, Problem 3–52 shows that it is necessary
and sufficient that
det(
A

λI
) = 0
.
(Otherwise Cramer’s rule yields
x
= 0.) This equation is quite interesting. The quantity
det
a
11

λ
a
12
. . .
a
1
n
a
21
a
22

λ
. . .
a
2
n
.
.
.
a
n
1
a
n
2
. . .
a
nn

λ
can in principle be written out in detail, and it is then seen that it is a polynomial in
λ
of
degree
n
. This polynomial is called the
characteristic polynomial
of
A
; perhaps it would be
more consistent to call it the
eigen
polynomial, but no one seems to do this.
The only term in the expansion of the determinant which contains
n
factors involving
λ
is
the product
(
a
11

λ
)(
a
22

λ
)
. . .
(
a
nn

λ
)
.
Thus the coefficient of
λ
n
in the characteristic polynomial is (

1)
n
. In fact, that product is
also the only term which contains as many as
n

1 factors involving
λ
, so the coefficient of
λ
n

1
is (

1)
n

1
(
a
11
+
a
22
+
· · ·
+
a
nn
). This introduces us to an important number associated
with the matrix
A
, called the
trace
of
A
:
trace
A
=
a
11
+
a
22
+
· · ·
+
a
nn
.
Notice also that the polynomial det(
A

λI
) evaluated at
λ
= 0 is just det
A
, so this is the
constant term of the characteristic polynomial. In summary,
det(
A

λI
) = (

1)
n
λ
n
+ (

1)
n

1
(trace
A
)
λ
n

1
+
· · ·
+ det
A.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Hatcher
 Calculus, Linear Algebra, Critical Point, Derivative, Matrices, Orthogonal matrix, symmetric matrices, real symmetric matrix

Click to edit the document details