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Math 415 Final Examination and Solutions - May 5, 2008 - R.
Muncaster
- (10 points)
(a) Define what it means for vectors v 1 ,... , vn to be linearly independent. Be
precise.
Solution: (4 points) Vectors v 1 ,... , vn are linearly independent if the only
constants c 1 ,... , cn for which
c 1 v 1 + · · · + cnvn = 0
are c 1 = 0,... , cn = 0.
(b) Prove that if the vectors v 1 ,... , vn are mutually orthogonal, then they are
also linearly independent.
Solution: (6 points) Let
c 1 v 1 + · · · + cnvn = 0
Take the inner product of both sides with v 1 to get
c 1 v 1 · v 1 + c 2 v 2 · v 1 + · · · + cnvn · v 1 = 0
so c 1 ‖v 1 ‖ + c 2 0 + · · · + cn0 = 0
or c 1 ‖v 1 ‖ = 0
Assuming all the vectors are non-zero (one bonus mark for mentioning
this), we conclude that c 1 = 0. In the same way we can show that all the
c’s are zero
- (10 points) Let
A =
(a) Find a basis for the cokernel of A. What is the rank of A?
Solution: (points 6)
A
T
so y = z, x = −y − 2 z = − 3 z. Thus
x
y
z
− 3 z
z
z
(^) = z
and so
)T
forms a basis of the cokernel.
(b) Find all values of the constant a for which the system
x
y
z
w
a
has a solution and what is the dimension of the solution set?
Solution: (4 points) By Fredholm’s Criterion, this system will have a so-
lution if the right-hand side is orthogonal to the cokernel, i.e. if
a 3 4
= − 3 a + 3 + 4 = 0, i.e. a =
The calculation in part a) shows that A has two linearly independent rows,
i.e. rank A is 2, so there are two linearly independent columns too, i.e. 2
pivots. This means there are two free variables and so the solution set is
2 dimensional.
- (15 points) Consider the electical network shown below. Wires 1, 2, 4 and 5
have resistance 1 ohm and wire 3 has resistance 2 ohms and there is a 10 volt
battery in wire 1. Find the general form of the voltage potentials at all nodes
(i.e. do not assume that one of the nodes is grounded).
Solution: The incident matrix, conductance matrix, and battery vector here are
A =
, C =
1 2
, b =
turn the vectors
w 1 =
(^) , w 2 =
(^) , w 3 =
into an orthonormal set.
Solution: First set v 1 = w 1. Then we compute
‖v 1 ‖
2
T
〈v 1 , w 2 〉 =
T
Thus we can take v 2 = w 2 since v 1 and w 2 are already orthogonal in this inner
product. Now for some more calculations:
‖v 2 ‖
2
T
〈v 1 , w 3 〉 =
T
〈v 2 , w 3 〉 =
T
Thus we define v 3 via Gram-Schmidt:
v 3 = w 3 −
〈v 1 , w 3 〉
‖v 1 ‖
2
v 1 −
〈v 2 , w 3 〉
‖v 2 ‖
2
v 2
3 7 3 14 0
and note that
‖v 3 ‖
2
3 7 3 14
0
T
3 7 3 14
0
Now all we need to do is normalize:
u 1 =
√^1 12 √^2 12 3 √ 12
, u 2 =
√^4 42 √^5 42
0
(^) , u 3 =
√^2 14 √^1 14
0
- (15 points) Solve the initial value problem
x˙ = 3x + y − z, x(0) = 1
y˙ = x + 3y − z, y(0) = 2
z˙ = 3x + 3y − z, z(0) = − 1
(Hint: one eigenvalue is λ = 1 and another is λ = 2, so you just need to find
the third one.).
Solution: We need to solve the eigenvalue problem first:
det
3 − λ 1 − 1
1 3 − λ − 1
3 3 − 1 − λ
(^) = − 8 λ + 5λ^2 − λ^3 + 4
= −(λ − 1)(λ − 2)
2
so λ = 1 and λ = 2 (double).
Case λ = 1:
(A − I)v =
a
b
c
2 a + b − c
a + 2b − c
3 a + 3b − 2 c
Subtract the first two to get a − b = 0, so b = a. Then the first gives c =
2 a + b = 3a. Thus we can use v =
)T
Case λ = 2:
(A − 2 I)v =
a
b
c
a + b − c
a + b − c
3 a + 3b − 3 c
Thus c = a + b so
v =
a
b
a + b
(^) = a
(^) + b
The general solution is then
x(t) = c 1 e
t
(^) + c 2 e
2 t
(^) + c 3 e
2 t
c 1 e
t
2 t
c 1 e
t
2 t
3 c 1 e
t
2 t
2 t
The initial condition then gives us
(^) = x(0) =
c 1 + c 2
c 1 + c 3
3 c 1 + c 2 + c 3
These two are both orthogonal to the first eigenvector, but are not orthog-
onal to each other, so we need a little Gram-Schmidt:
1 2 1
1 2
gives an alternate to the third vector orthogonal to the second. Now just
normaliz to get
Q =
√^1 3
√^1 2
√^1 6 √^1 3
√^2 6 √^1 3
√^1 2
√^1 6
,^ Λ =
- (10 points) Find the closest point on the hyperplane x + y + z + w = 0 to the
vector b =
)T
Solution: First find a basis for the hyperplane: x = −y − z − w so
x
y
z
w
−y − z − w
y
z
w
= y
Now we set
A =
, K = A
T A =
(^) , f = AT^ b =
and proceed to solve the normal equations Kx = j :
3 2
1 2
1 2
3 2
3 2
1 2
4 3
So x 3 = −
3 4
, x 2 = −
1 3
x 3 =
1 4
, x 1 =
1 2
3 4
1 4
3 4
and therefore the
closest point is
- (10 points) Consider the vector space P
(2)
. On this space consider the function
(with v(x) = ax
2
L[v] = L[ax
2
2
(a) Verify that L is linear
Solution: (5 points) Set v 1 = a 1 x
2
- b 1 x + c 1 and v 2 = a 2 x
2
Then
L[cv 1 + dv 2 ] = L[(ca 1 + da 2 )x
2
- (cb 1 + db 2 )x + (cc 1 + dc 2 )]
= (cb 1 + db 2 )x
2
- (cc 1 + dc 2 )x + (ca 1 + da 2 )
= c(b 1 x
2
2
= cL[v 1 ] + dL[v 2 ]
(b) Relative to the basis p 1 (x) = 1, p 2 (x) = x, p 3 (x) = x
2 , what is the matrix
A that represents L?
Solution: (5 points)
L[p 1 ] = L[1] = x = 0p 1 (x) + 1p 2 (x) + 0p 3 (x)
L[p 2 ] = L[x] = x
2 = 0p 1 (x) + 0p 2 (x) + 1p 3 (x)
L[p 3 ] = L[x
2 ] = 1 = 1p 1 (x) + 0p 2 (x) + 0p 3 (x)
so
A =
