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Math 415 Exam # 2 Solutions
- (10 points)
(a) Demonstrate why
〈f, g〉 =
− 1
f (x)g(x)xdx
is not an inner product on F([− 1 , 1]).
Solution: (5 points) Show that positivity fails. Take f (x) = 1, for example.
Then
2 = 〈 1 , 1 〉 =
− 1
1 · 1 · xdx =
x
2 |
1 − 1 = 0
but f (x) = 1 is not the zero vector.
(b) Find a weighted inner product 〈v, w〉 = av 1 w 1 + bv 2 w 2 + cv 3 w 3 for which
the vectors v =
)T
and w =
)T
are orthogonal.
Solution: (5 points) The inner product of the two vectors is 〈v, w〉 =
a · 1 · 0 + b · 2 · (−1) + c · (−1) · 3 = − 2 b − 3 c. For this to be zero, b
and c would have to be of opposite signs. But for a, b, and c to define a
weighted inner product, they all need to be positive. So this problem has
no solution.
- (15 points)
(a) Find the A = LDL
T factorization of
A =
Solution: (6 points) By Gaussian-elimination
and so
D =
(^) and L =
(b) Given that
K =
write the quadratic form q(x) = x
T Kx as a sum of squares of expressions
in x. Is q(x) positive definite, negative definite, or indefinite, and why?
Solution: (9 points) Set
y = L
T x =
x 1
x 2
x 3
x 1 + 5x 2 + 6x 3
x 2 + 7x 3
x 3
Then q(x) = x
T Kx = y
T Dy = 9y
2 1 −^2 y
2 2 + 7y
2 3 = 9(x^1 + 5x^2 + 6x^3 )
2 −
2(x 2 + 7x 3 )
2
2 3.^ We see that^ q(x) is indefinite because the diagonal
elements of D are not all positive and not all negative.
- (10 points)
(a) Define what is meant by an orthonormal basis. Give an example of one in
R
2 relative to the dot product that does NOT contain either of the stan-
dard basis vectors.
Solution: (5 points) An orthonormal basis is a basis that is mutually or-
thogonal and in which all vectors are unit vectors. An example is
v 1 =
, v 2 =
(b) Give a proof that if v 1 , v 2 , and v 3 are mutually orthogonal, then they are
linearly independent.
Solution: (5 points) If
c 1 v 1 + c 2 v 2 + c 3 v 3 = 0
then taking inner products of both sides with v 1 gives us
0 = c 1 v 1 · v 1 + c 2 v 2 · v 1 + c 3 v 3 · v 1 = c 1 ||v 1 ||
2
- c 2 0 + c 3 0 = c 1 ||v 1 ||
2
Since v 1 is none-zero, we see that c 1 = 0. In the same way you can show
that c 2 and c 3 are zero.
- (10 points) Using distance measured by the dot product, find the point in the
plane W spanned by v 1 and v 2 closest to the vector v where
v 1 =
(^) , v 2 =
(^) , v =
What is the distance between W and v?
Solution: You can either switch to an orthogonal basis and do an orthogonal
projection or write down and solve the normal equations.
Method 1: Set
w 1 = v 1 =
(^) , w 2 =^ v 2 −^
v 2 · w 1
||w 1 ||^2
w 1 =
1 2 1 1 2
is connected. Thus the kernel of A
T is the vector of all 1’s, and so the
compatibility condition is
b 1
b 2
b 3
b 4
b 5
= b 1 + b 2 + b 3 + b 4 + b 5
(c) One solution of the system below is x =
)T
. What
equations should be added to the system to find the unique solution of
minimum norm?.
1 − 1 0 0 0
1 0 − 1 0 0
x 1
x 2
x 3
x 4
x 5
Solution: (3 points) The coefficient matrix here is the incident matrix for a
connected digraph, so its kernel is spanned by the vector of all ones. Thus,
to find the solution of minimum norm we must add the equation
x 1
x 2
x 3
x 4
x 5
= x 1 + x 2 + x 3 + x 4 + x 5
