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Unformatted text preview: Math 235 Assignment 8 Practice Problems Solutions 1. Sketch the graph of the equation x2 − 4x1 x2 + x2 = 8. Show both the original axes and
1
2
the new axes. Find the equation of the asymptotes.
Solution: The corresponding symmetric matrix is 1 −2
. We ﬁnd that the character−2 1 1 − λ −2
= λ2 − 2λ − 3 = (λ − 3)(λ + 1). So, we have
−2 1 − λ
−2 −2
11
eigenvalues λ1 = 3 and λ2 = −1. For λ1 = 3 we get A − λ1 I =
∼
.
−2 −2
00
2 −2
1
Thus, a corresponding eigenvector is v1 =
. For λ2 we get A − λ2 I =
∼
−1
−2 2
1 −1
.
00
1
2
2
A corresponding eigenvector is v2 =
. Thus, we have the hyperbola 3y1 − y2 = 8 with
1
principal axis v1 for y1 and v2 for y2 .
√
2
2
The asymptotes of the hyperbola are when 0 = 3y1 − y2 . Thus, they are y2 = ± 3y1 . We
have (ﬁnding the inverse of the change of basis matrix by taking the transpose)
istic polynomial is C (λ) = 1 x1 − x2
1 1 −1 x1
y1
=√
=√
x2
y2
21 1
2 x1 + x2
So, the asymptotes for the hyperbola in the x1 x2 plane are
√
x1 + x2 = ± 3(x1 − x2 ) Solving for x2 gives Graphing gives: √
−1 ± 3
√ x1
x2 =
1± 3 2 2. Sketch the graph of 5x2 + 6xy − 3y 2 = 15 showing both the original and new axes. Find
the equation of the asymptotes.
Solution: The corresponding symmetric matrix is 53
. We ﬁnd that the characteristic
3 −3 5−λ
3
= λ2 − 2λ − 24 = (λ − 6)(λ + 4). So, we have
3
−3 − λ
−1 3
1 −3
eigenvalues λ1 = 5 and λ2 = −4. For λ1 = 6 we get A − λ1 I =
∼
.
3 −9
00
3
93
1 1/3
Thus, a corresponding eigenvector is v1 =
. For λ2 we get A − λ2 I =
∼
.
1
31
00
−1
2
A corresponding eigenvector is v2 =
. Thus, we have the hyperbola 6x2 − 4y1 = 15
1
3
with principal axis v1 for x1 and v2 for y1 .
The asymptotes of the hyperbola are when
√
√
2
0 = 6x2 − 4y1 = ( 6x1 + 2y1)( 6x1 − 2y1)
1
polynomial is C (λ) = We have (ﬁnding the inverse of the change of basis matrix by taking the transpose)
1 3x + y
1
31x
x1
.
=√
=√
−1 3 y
y1
10
10 x − 3y
√
√
√
√
So, the asymptotes are 0 = 6x1 ± 2y1 = 6(3x + y ) ± 2(x − 3y ) = (3 6 ± 2)x + ( 6 ∓ 6)y,
√
hence y = (3 √6±2) x ⇒ y ≈ 2.633x, y ≈ −0.633x.
− 6±6
Graphing gives: 3 3. Sketch the graph of 6x2 + 4xy + 3y 2 = 14 showing both the original and new axes.
Solution: The corresponding symmetric matrix is 62
. We ﬁnd that the characteristic
23 polynomial is
C (λ) = 6−λ
2
= λ2 − 9λ + 14 = (λ − 2)(λ − 7).
2
3−λ So, we have eigenvalues λ1 = 2 and λ2 = 7. For λ1 = 2 we get
A − λ1 I =
Thus, a corresponding eigenvector is v1 =
A − λ1 I = 42
21
∼
.
21
00
1
. For λ1 = 7 we get
−2 −1 2
−1 2
∼
.
2 −4
00 Thus, a corresponding eigenvector is v2 = 2
. Thus, we have the ellipse
1 2x2 + 7x2 = 14
1
2
with principal axis v1 for x1 and v2 for y1 . Graphing gives: 4 4. For each quadratic form Q(x), determine the corresponding symmetric matrix A. By
diagonalizing A, express Q(x) in diagonal form and give an orthogonal matrix that diagonalizes A. Classify each quadratic form.
(a) Q(x, y, z ) = 7x2 + 8xy + y 2 + 8xz − 16yz + z 2 . 74
4
Solution: We have A = 4 1 −8 so the characteristic polynomial is
4 −8 1
7−λ
4
0
7−λ
4
0
7−λ
4
4
4
1 − λ −9 + λ
4
1 − λ −9 + λ =
4
1 − λ −8 =
C (λ) =
4
−7 − λ
0
4
−8
9−λ
4
−8 1 − λ
= −(λ − 9)(λ2 − 81) = −(λ − 9)2 (λ + 9). Hence, the eigenvalues are λ = 9 and λ = −9. For λ = 9 we get −2 4
4
1 −2 −2
0 .
A − λI = 4 −8 −8 ∼ 0 0
4 −8 −8
00
0 2
2
Thus, we get eigenvectors v1 = 1 and v2 = 0. Since we need orthogonal vectors, we
0
1
apply the GramSchmidt procedure. We take w1 = v1 and 2/5
2
2
4
v2 · w1
w1 = 0 − 1 = −4/5 .
w2 = v2 −
w1 2
50
1
1
For λ = −9 we get 16 4
4
1 0 1/2
A − λI = 4 10 −8 ∼ 0 1 −1 .
4 −8 10
00 0 −1/2
Thus, we get eigenvector v3 = 1 .
1 Normalizing the vectors, we get that the orthogonal
√
√
2/√5 2/ √
45
1/ 5 −4/ 45
P=
√
0
5/ 45 x1
x
2
2
2 y1 = P T y .
and Q = 9x1 + 9y1 − 9z1 , where
z1
z matrix which diagonalizes A is −1/3
2/3 ,
2/3 Since A has both positive and negative eigenvalues it is indeﬁnite. 5 (b) Q(x) = x2 + 3x1 x2 + x2 .
1
2
Solution: We have A =
We have
C (λ) = 1 3/2
1 − λ 3/2
so A − λI =
.
3/2 1
3/2 1 − λ
1
5
5
1 − λ 3/2
= λ2 − 2λ − = (λ + )(λ − ).
3/2 1 − λ
4
2
2 1
5
Thus, the eigenvalues of A are λ1 = − 2 and λ2 = 2 .
1
For λ1 = − 2 we get A − λ1 I = 3/2 3/2
11
∼
, so v1 =
3/2 3/2
00 1
√
2 −1
.
1 −3/2 3/2
1 −1
1
1
∼
, so v2 = √2
.
3/2 −3/2
00
1
√
√
−1/ 2 1/√2
−1/2 0
√
Hence, we can take P =
to get D =
.
0
5/2
1/ 2 1/ 2 For λ2 = 5
2 we get A − λ2 I = Then, using the change of variables
√
√
−1/ 2 1/√2
√
x = Py =
1/ 2 1/ 2 √
√
y1
−(1/ 2)y1 + (1/ 2)y2
√
√
=
,
y2
(1/ 2)y1 + (1/ 2)y2 12
2
we get Q(x) = − 2 y1 + 5 y2 .
2 Since one of the eigenvalues of A are positive and the other is negative, it follows that
Q(x, y ) is indeﬁnite.
(c) Q(x) = 4x2 + 4x1 x2 + 4x1 x3 + 4x2 + 4x2 x3 + 4x2 .
1
2
3 422
Solution: We have A = 2 4 2 so the characteristic polynomial is
224
C (λ) = 4−λ
0
2
4−λ
0
2
4−λ
2
2
2
2−λ
2
2
2−λ
2
2
4−λ
2
=
=
4
0
6−λ
2
−(2 − λ) 4 − λ
2
2
4−λ = −(λ − 2)(λ2 − 10λ + 16) = −(λ − 2)2 (λ − 8).
Hence, the eigenvalues are λ1 = 2 and λ2 = 8.
For λ1 = 2 we get 222
111
A − λ1 I = 2 2 2 ∼ 0 0 0 .
222
000 6 −1
−1
Thus, we get eigenvectors v1 = 1 and v2 = 0 . Since we need orthogonal vectors,
0
1
we apply the GramSchmidt procedure. We take w1 = v1 and −1
−1/2
−1
1
v2 · w1
w1 = 0 − 1 = −1/2 .
w2 = v2 −
w1 2
20
1
1
For λ2 = 8 we get −4 2
2
1 0 −1
A − λ2 I = 2 −4 2 ∼ 0 1 −1 .
2
2 −4
00 0 1
1.
Thus, we get eigenvector v3 =
1
Normalizing the vectors, we get that the orthogonal
√
√ −1/ 2 −1/√6
√
P = 1/ 2 −1/ 6
√
0
2/ 6 y1
x1
2
2
2
y2 = P T x2 .
and Q = 2y1 + 2y2 + 8y3 , where
y3
x3 matrix which diagonalizes A is
√
1 / √3
1 / √3 ,
1/ 3 Since all the eigenvalues of A are positive, A is positive deﬁnite.
5. Let A be a positive deﬁnite symmetric matrix. Prove that:
a) the diagonal entries of A are all positive.
Solution: If A is a positive deﬁnite symmetric matrix, then xT Ax > 0 for all x = 0. Hence,
for 1 ≤ i ≤ n, we have
aii = eT Aei > 0
i
as required.
b) A is invertible.
Solution: If A is a positive deﬁnite symmetric matrix, then all the eigenvalues of A are
positive. Thus, since the determinant of A is the product of the eigenvalues, we have that
det A > 0 and hence A is invertible. 7 c) P T AP is positive deﬁnite for any orthogonal matrix P .
Solution: Pick any y ∈ Rn , y = 0. Then, y T P T AP y = (P y )T A(P y) > 0 since xT Ax > 0 for all x = 0 as A is positive deﬁnite. Hence, P T AP is positive deﬁnite.
6. Let A and B be symmetric n × n matrices whose eigenvalues are all positive. Show that
the eigenvalues of A + B are all positive.
Solution: We ﬁrst must observe that A + B is symmetric. Since the eigenvalues of A and
B are all positive, the quadratic forms xT Ax and xT Bx are positive deﬁnite. Let x = 0.
Then xT Ax > 0 and xT Bx > 0 , so xT (A + B )x = xT Ax + xT Bx > 0 , and the quadratic
form xT (A + B )x is positive deﬁnite. Thus the eigenvalues of A + B must be positive. 1 −1
7. Let A = 1 −4. Find the maximum and minimum value of Ax for x ∈ R2 subject
−4 1
to the constraint x = 1.
18 −9
. Hence, the eigenvalues of AT A are λ1 = 27 and
−9 18
√
λ2
√ = 9. Thus the maximum of Ax subject to x = 1 is 27 and the minimum is
9 = 3.
Solution: We have AT A = ...
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This note was uploaded on 01/21/2012 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN
 Asymptotes

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