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Math 415 - Assignment 5 Solutions
Problems: 3.1.1, 3.1.3, 3.1.5, 3.1.14, 3.1.20, 3.2.37, 3.4.1, 3.4.2, 3.4.6, 3.4.22(vii), 3.4.24 (vii),
3.5.3, 3.5.6 (c)
Problem 3.1.
Symmetry: 〈v, w〉 = v 1 w 1 − v 1 w 2 − v 2 w 1 + bv 2 w 2 = w 1 v 1 − w 1 v 2 − w 2 v 1 + bw 2 v 2 = 〈w, v〉
Linearity:
〈cv+du, w〉 = (cv 1 + du 1 )w 1 − (cv 1 + du 1 )w 2 − (cv 2 + du 2 )w 1 + b(cv 2 + du 2 )w 2
= cv 1 w 1 + du 1 w 1 − cv 1 w 2 − du 1 w 2 − cv 2 w 1 − du 2 w 1 + bcv 2 w 2 + bdu 2 w 2
= c(v 1 w 1 − v 1 w 2 − v 2 w 1 + bv 2 w 2 ) + d(u 1 w 1 − u 1 w 2 − u 2 w 1 + bu 2 w 2 )
= c 〈v, w〉 + d 〈u, w〉
Linearity in the second argument now follows from this calculation and symmetry.
Positivity: 〈v, v〉 = v 2 1 − v 1 v 2 − v 2 v 1 + bv 2 2 = v 2 1 − 2 v 1 v 2 + v 2 2 − v 2 2
- bv 2 2 = (v 1 − v 2 ) 2
- (b − 1)v 2 2
This expression is a sum of squares with positive coefficients when b > 1. Thus 〈v, v〉 ≥ 0 and
vanishes only if v 2 = 0 and v 1 = v 2 i.e. v = 0. On the other hand, if b ≤ 1, then any vector with
v 1 = v 2 6 = 0 will make 〈v, v〉 either zero or negative.
Problem 3.1.
Check positivity: 〈v, v〉 = v
2 1 +^ v^1 v^2 +^ v^2 v^1 +^ v
2 2 = (v^1 +^ v^2 )
2
. This shows that the vector v = (
1 − 1
)T
is non-zero but 〈v, v〉 = 0. Thus positivity is violated.
Problem 3.1.
(a) Euclidean: v 2 1
- v 2 2 = 1. This is a circle of radius 1 in the v 1 vs v 2 plane.
(b) Here 2v
2 1
2 2 = 1. This is an ellipse in the v 1 vs v 2 plane of imnor axis 1/
5 along the v 2 axis
and major axis 1/
2 along the v 1 axis.
(c) Here v
2 1 −^2 v^1 v^2 + 4v
2 2 = (v^1 −^ v^2 )
2
2 2 =^ y
2 1 + 3y
2 2 = 1 where^ y^1 =^ v^1 −^ v^2 and^ y^2 =^ v^2. This
an ellipse in the y 1 vs y 2 plane, a rotation of the v 1 vs v 2 plane.
(d) See b) and c).
Problem 3.1.
Using the dot product definition as v · w = v
T w, we find that
v · (Aw) = v
T Aw
= (A
T v)
T w
= (A
T v) · w
Problem 3.1.
(a)
x, 1 + x 2
1 0
x(1 + x 2 )dx =
3 4
, ‖x‖ =
1 0
x 2 dx =
1 3
∥1 + x^2
1 0
(1 + x 2 ) 2 dx =
2 15
(b)
x, 1 + x
2
− 1 x(1 + x
2 )dx = 0, ‖x‖ =
− 1 x 2 dx =
1 3
1 + x
2
− 1 (1 + x 2 ) 2 dx =
2 15
(c)
x, 1 + x
2
0
x(1 + x
2 )xdx =
8 15 , ‖x‖ =
0
x 2 xdx =
1 2
1 + x
2
0
(1 + x 2 ) 2 xdx =
1 6
Problem 3.2.
(a) Cauchy Schwarz Inequality:
1 0
f (x)g(x)e x dx
1 0
f (x) 2 e x dx
1 0
g(x) 2 e x dx
Triangle Inequality:
1 0
(f (x) + g(x)) 2 e x dx ≤
1 0
f (x) 2 e x dx +
1 0
g(x) 2 e x dx
(b)
1 0
f (x)g(x)e x dx =
1 0
e 2 x dx =
1 2
e 2 −
1 2
1 0
f (x) 2 e x dx =
1 0
e x dx = e − 1 ,
1 0
g(x) 2 e x dx = ∫ (^1)
0
e
3 x dx =
1 3 e
3 −
1 3
0
(f (x)+g(x))
2 e
x dx =
0
(1+e
x )
2 e
x dx =
0
(e
x +2e
2 x +e
3 x )dx = e+e
2
1 3 e
3 −
7 3
So Cauchy Schwarz:
2 e
2 −
1 2
e − 1
1 3 e 3 −
1 3 = 3.3063 - verified!
and Triangle:
e + e 2
1 3
e 3 −
7 3
e − 1 +
1 3
e 3 −
1 3
= 3.8331 - verified!
(c) The angle between f (x) = 1 and g(x) = e
x is θ where
cos θ =
〈f,g〉 ‖f ‖‖g‖
1 2 e 2 − 1 2 √ e− 1
1 3 e^3 − 1 3
=. 9662 ⇒ θ = cos
− 1 (.9662) = .26074 radians =. 26074
180 π
degrees
Problem 3.4.
(a) Yes: Diagonal with all positive entries
(b) No: Associated quadratic form is q(x) = q(x 1 , x 2 ) = 3x 1 x 2 which is negative any time that
x 1 < 0 and x 2 > 0.
(c) No: The (1,1) entry is positive but the determinant is 1 × 1 − 2 × 2 = −3 is negative.
(d) No: The (1,1) entry is positive but the determinant is 5 × (−2) − 3 × 3 = −19 is negative.
(e) Yes: The (1,1) entry is positive and the determinant is 1 × 3 − (−1) × (−1) = 2 is positive.
(f) No: The matrix is not symmetric and so cannot be positive definite.
Problem 3.4.
q(x) =
x 1 x 2
x 1
x 2
= (x 1 + 2x 2 ) x 1 + (2x 1 + 3x 2 ) x 2 = x 2 1
x
2 1 + 4x^1 x^2 + 4x
2 2 −^ x
2 2 = (x^1 + 2x^2 )
2 − x
2
- We see now that^ q(1,^ 0) = 1^ >^ 0 and^ q(−^2 ,^ 1) =^ −^1 <^ 0.
Problem 3.4.
The quadratic form of K is q(x) = x
T Kx and we know this is always positive unless x = 0. The
quadratic form of cK is ˆq(x) = x
T (cK)x = c(x
T Kx) = cq(x). Given that c > 0 we conclude that
qˆ(x) is positive unless x = 0.
Problem 3.4.22(vii)
(a) K =
(b) By Gaussian Elimination:
Since there are three pivots, the vectors are linearly independent, and so K is positive definite.
(c) Nothing to do! kerK = { 0 }
positive definite iff q(x, y, z) is positive definite, iff ˆq(r, s, t) is positive define (which is true since it
is a sum of squares with positive coefficients).
Problem 3.5.6 (c)
q(x 1 , x 2 , x 3 ) = 2x
2 1
2 2 − 2 x 2 x 3 + 2x
2 3
= 2(x
2 1 + 2x^1 (^
x 2 − 2 x 3
)) + 2x
2 2 −^2 x^2 x^3 + 2x
2 3
= 2(x
2 1 + 2x^1 (^
x 2 − 2 x 3
x 2 − 2 x 3
2 ) − 2(
x 2 − 2 x 3
2
2 2 −^2 x^2 x^3 + 2x
2 3
= 2(x 1 +
x 2 − 2 x 3
2 −
x
2 2 +
x 2 x 3 −
x
2 3 + 2x
2 2 −^2 x^2 x^3 + 2x
2 3
= 2(x 1 +
x 2 − 2 x 3
2
x
2 2 −^
x 2 x 3 +
x
2 3
= 2(x 1 +
x 2 − 2 x 3
2
(x
2 2 −^2 x^2 (
x 3 )) +
x
2 3
= 2(x 1 +
x 2 − 2 x 3
2
(x
2 2 −^2 x^2 (
x 3 ) + (
x 3 )
2 ) −
x 3 )
2
x
2 3
= 2(x 1 +
x 2 − 2 x 3
2
(x 2 −
x 3 )
2
x
2 3
which is a sum of squares with positive coefficients. Moreover, it vanishes only if x 3 = 0, x 2 =
2 5 x 3 =
0 , x 1 = −
1 4 (x 2 − 2 x 3 ) = 0. Done!
