(l) Orthogonal changes of coordinates
x
=
U
x
′
affect the symmetric matrix
Q
of a quadratic form by similarity transformations
U
t
QU
=
U
−
1
QU
and hence
preserve the characteristic polynomial of
Q
. According to Orthogonal Diagonaliza
tion Theorem the quadratic form can be transformed by such changes of coordi
nates to one of the normal forms
AX
2
1
+
CX
2
2
with the characteristic polynomial
(
λ
−
A
)(
λ
−
C
).
Thus the coefficients
A,C
in the normal form are roots of the
characteristic polynomial
vextendsingle
vextendsingle
vextendsingle
vextendsingle
λ
−
a
−
b
−
b λ
−
c
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
λ
2
−
(
a
+
c
)
λ
+ (
ac
−
b
2
)
.
30
1. GEOMETRY ON THE PLANE
We complete this section with a proof of Orthogonal Diagonalization Theorem.
It has several reformulations (see Principal Axes Theorem in 1
.
2
.
2, Example 1
.
3
.
3(
a
)
and Section 1
.
3
.
4(
IV
)) equivalent to the following one:
For any symmetric matrix
Q
there exists an orthogonal matrix
U
such that
U
−
1
QU
is diagonal.
Proof.
The discriminant of the characteristic polynomial of
Q
equals
(
a
+
c
)
2
−
4(
ac
−
b
2
) =
a
2
+ 2
ac
+
c
2
−
4
ac
+ 4
b
2
= (
a
−
c
)
2
+ 4
b
2
and is nonnegative. This means that the characteristic polynomial of a symmetric
matrix has a real root
λ
1
and hence — a real eigenvector, a nontrivial solution to
the homogeneous linear system
Q
v
1
=
λ
1
v
1
. Let
v
2
be a nonzero vector orthogonal
to
v
1
. Then
(
Q
v
2
,
v
1
)
=
(
v
2
,Q
v
1
)
=
(
v
2
,λ
1
v
1
)
=
λ
1
(
v
2
,
v
1
)
= 0
.
Thus
Q
v
2
is also orthogonal to
v
1
and is therefore proportional to
v
2
:
Q
v
2
=
λ
2
v
2
with some real coefficient
λ
2
. Normalizing the perpendicular eigenvectors
v
1
and
v
2
to the unit length we get an orthonormal basis
u
1
=
v
1
/

v
1

,
u
2
=
v
2
/

v
2

of
eigenvectors. Let
U
denote the matrix whose columns represent
u
1
and
u
2
. It is
orthogonal since
U
t
U
=
bracketleftbigg
u
t
1
u
1
u
t
1
u
2
u
t
2
u
1
u
t
2
u
2
bracketrightbigg
=
bracketleftbigg
1
0
0
1
bracketrightbigg
.
The columns of
QU
represent
λ
1
u
1
and
λ
2
u
2
.
Thus
QU
=
U
Λ, where Λ =
bracketleftbigg
λ
1
0
0
λ
2
bracketrightbigg
is the diagonal matrix of eigenvalues, and hence
U
−
1
QU
= Λ.
Exercises
1
.
5
.
2
.
(a) Verify the multiplicative property of determinants.
(b) Give an example of two matrices which are not similar but have the same characteristic
polynomial.
(c) Show that tr(
A
+
B
) = tr
A
+ tr
B
, tr(
AB
−
BA
) = 0.
(d) Find the intersection point of the lines 3
x
1
+ 2
x
2
= 1 and
x
1
+ 2
x
2
= 3.
Find the
coordinates of the intersection point in the basis
f
1
= (1
,
3)
,
f
2
= (3
,
1).
(e) Find out which of the following quadratic curves is an ellipse and which is a hyperbola
and compute their semiaxes (that is the coefficients
a
1
,a
2
in the normal forms
x
2
1
/a
2
1
±
x
2
2
/a
2
2
= 1):
2
x
2
1
+ 3
x
1
x
2
+
x
2
2
= 1
,
3
x
2
1
−
3
x
1
x
2
+
x
2
2
= 1
.
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 Spring '08
 Chorin
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Vector Space, linear transformation, HU