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Graphs of
Quadratic
Forms
Graphs of Quadratic Forms
In
JR
...
2
=
k,
and k is a constant
...
However, by applying diagonalization to the problem of determining
these graphs, we see a very clear interpretation of eigenvectors
...
Moreover, this approach to
these graphs leads to a
the form
Q(x)
classification
= JR
...
Classification is a
useful process because it allows us to say "I really only need to understand these few
standard cases
...
Observe that in general it is difficult to identify the shape of the graph of
bxJX2 + ex� =
axi +
k
...
However, it is
easy to sketch the graph of a
c�
k
...
Of course, we first need
xi + =
=
to determine how diagonalizing the quadratic form will affect the graph
...
Then an orthogonal
1, which diagonalizes Q(x), corresponds to a rotation in
with det
Proof: Let A
where
...
2
there exists an orthonormal basis {v, w} of
w = [Ww2J]
_,
...
Let v
...
2
...
Therefore, there exists an angle e
[]
w=+
[= �
]
sine
...
We choose
have
p
sine
_,
cose
c se
- sine
sme
cose
so that det
P=
1
...
Finally, from our work in Section 8
...
= [��� ]
-
ne
e
...
we would find that
P
corresponds to a rotation and a
In practice, we do not need to calculate the angle of rotation
...
In particular, taking y =
and taking y =
[�]
[�l
we get
gives
That is, the new y1-axis corresponds to the vector v1 in the x1x2-plane, and the y2-axis
corresponds the vector V2 in the x1x2-plane
...
EXAMPLE 1
i
Sketch the graph of the equation 3x + 4x1x2 = 16
...
A =
matnx
[ )
3
2
+ 4x1x2 corresponds to the symmetric
2
...
1 1s
...
Thus, by an orthogonal change of
coordinates, the equation can be brought into the diagonal form:
f
�
4y -y =l6
This is an equation of a hyperbola, and we can sketch the graph in the y1y2-plane
...
The asymptotes of the hyperbola are determined by the equation 4y -Y =
0
...
With this information, we obtain the graph in Figure 8
...
1
...
Hence, we need to find the
eigenvectors of A
...
3
...
=
16,
Y1-axis
EXAMPLE 1
(continued)
Thus, a basis for the eigenspace is {v2}, where v2
but
[-�]
[ �]
-
=
[ �l
...
)
Now we sketch the graph
T
of 3x + 4x1x2
16
...
(For clarity, in Figure 8
...
2 we
[�]
have shown the vector
stead of
[�]
in-
...
Then, relative to these
new axes, we sketch the graph
f
�
of the hyperbola4y -y
=
16
...
3
...
In order to include the
asymptotes in the sketch, we
rewrite their equations in stan
Figure 8
...
2
dard coordinates
...
onal change of coordinates
matrix in this case is given by
P
(This is a rotation of the axes through angle e;: 0
...
) Thus,
=
}s
[� �]
...
This gives
and
Then one asymptote is
The other asymptote is
Thus, in standard coordinates, the asymptotes are 3x1 + 4x2
=
0 and x1
=
0
...
[� �l
Solution: The corresponding symmetric matrix is
;!1
=
2 and ;!2
=
The eigenvalues are
7
...
) and (0,
-
(-fl, 0)
and
(--fl, 0)
and
-fi
...
3
...
In Fig
ure 8
...
4, the new Yt - and y2-axes determined by the eigenvectors are shown rela
tive to the standard axes, and the ellipse from Figure 8
...
3 is rotated into place
...
=
14 in the Y1Y2-plane
...
3
...
3
...
Since diagonalizing a quadratic form corresponds to a rotation, to classify all the
graphs of equations of the form
the form
A
...
2y� = k
...
1 and A
...
The distinct possibilities are displayed in Table 8
...
1
...
3
...
1 > 0, A
...
1> 0, A
...
1 > 0, A
...
1 = 0, A
...
1 < 0, A
...
The nondegenerate cases are the ellipses and hyperbolas,
which are conic sections
...
3 as the intersec
tion of a cone and a plane
...
However, the cases of parallel lines (in Table 8
...
1
...
It is also important to realize that one class of conic sections, parabolas, does not
appear in Table 8
...
1
...
2, the equation of a parabola is a quadratic equation,
but it
contains first-degree terms
...
The classification provided by Table 8
...
l suggests that it might be interesting
an equation of the form
to consider how degenerate cases arise as limiting cases of nondegenerate cases
...
3
...
xf + x� = 1 as A
...
...
--
...
...
...
·
----
Figure
8
...
5
= ±constant) arises from
...
...
...
/
·"" · ···
·· ··
Xt
1
...
= 1; as A
...
= 0, the graph is a pair of lines
...
xf
...
,
···
...
...
,
··
·
·
·
·
--
...
3
...
For example, the graph of xf - x� = - 1 is the same as the graph of
-xf + x� = 1, and this hyperbola may be obtained from the hyperbola xf - x� = 1 by
reflecting over the line x1 = x2 • However, for purposes of illustration, it is convenient
The graphs obtained for
k<
k > 0 and k = 0
...
Figure 8
...
6 shows that the case of intersecting lines
(k
0) separates the hyperbolas with intercepts on the x1 -axis Cxi
the hyperbolas with intercepts on the x -axis Cxi - 2x�
k < 0)
...
3
...
Diagonalize the quadratic form and sketch the graph of the equation
Show both the original axes and the new axes
...
=
k in JR3, there are similar results to what
we did above
...
The nondegenerate cases give ellipsoids, hyperboloids of one sheet, and
hyperboloids of two sheets
...
The usual standard form for the equation of an ellipsoid is
This is the case obtained by diagonalizing
Q(x)
=
x2
a�
+
x2
b�
x2
+
c;
=
1
...
In particular, if we write
An ellipsoid is shown in Figure 8
...
7
...
Figure 8
...
7
An ellipsoid in standard position
...
(a) k = 1; a hyperboloid of one sheet
...
3
...
(c) k = -1; a hyperboloid of two sheets
...
This form is obtained when k and two eigenvalues of the matrix of
x2
x2
a + b
�
;
-
x2
;
c
=
Q are positive
and the third eigenvalue is negative
...
Notice that if this is rewritten
1-
� + �=
¥i, it is clear that for every z there are values of x1 and x2 that satisfy the equation,
so that the surface is all one piece (or one sheet)
...
3
...
2
2
X2 X3
- 2=
The standard form of the equation for a hyperboloid of two sheets is 2 +
2
b
a
c
-1
...
Notice that if this is rewritten
� + � =-1 - ¥i, it is clear that for every
lx3I
S: -c
...
3
...
It is interesting to consider the family of surfaces obtained by varying k in the
x2 x2
x2
equation
+
� b;
a
-
;
c
=k, as in Figure 8
...
8
...
As k decreases towards -1, the waist has disappeared, and the graph is now a
hyperboloid of two sheets
...
3
...
13<0
hyperboloid of two sheets
cone
...
11,
...
3
...
The nondegenerate
cases are the ellipsoids and hyperboloids
...
c
a
b
x
z
z
z
(a)
(b)
(c)
Figure 8
...
9
i
�
Some degenerate quadric surfaces
...
x = 1 ,
i
�
parallel t o the x3-axis
...
x - x = 1 , parallel
i
�
to the x2-axis
...
x - x = 0
...
3
...
Note that paraboloidal sur
faces do not appear as graphs of the form Q(x) = k in R3 for the same reason that
2
parabolas do not appear in Table 8
...
1 for R : their equations contain first-degree
terms
...
Show both the original axes and the new axes
...
Show both the original axes and the
new axes
...
0
1
(d) A=
-2
15
...
AS For each of the following symmetric matrices, iden-
[1 ll
[� ]
[� -�]
1
12
...
A4 Sketch the graph of the equation Sxi +6x1x2-3x� =
[� n
[� -�]
0
-2
-1
-2
-2
0
8
(e) A=
1
-4
Homework Problems
�
i
Bl Sketch the graph of the equation 9x +4x1x2+6x =
90
...
i
�
B2 Sketch the graph of the equation x + 6x1x2 - 7 x =
32
...
i
B6 In each of the following cases, diagonalize the
quadratic form
...
(c)
� +4x1X2+x� = 8
...
f
= k for k = 1, 0, -1
...
B3 Sketch the graph of the equation x -4x1x2+x = 8
...
Show both the original axes and the new axes
...
Plot graphs of the original system
and of the diagonalized system