SOLUTION SET (PSET 2, 18.085)
Abstract.
Section 1.4:
Section 1.5:
Section 1.6:
All questions are taken from the book.
9,11,12
9,11,12
3,9,16,17,22,27
Solution 1. The distributed load f (x) = 1 is the integral of the loads
δ(x − a) at all points x = a. The free-fixed solution u(x) = 12 (1 − x2 ) is the
integral of the point-load solutions of section 1.3 in the book, i.e.,
(
1 − x , a ≤ x,
Q(x) =
(1)
1 − a , a ≥ x.
Thus we compute
Z x
Z 1
x2
1
1
u(x) =
(1 − x)da +
(1 − a)da = (1 − ) − (x − ) = (1 − x2 ). (2)
2
2
2
0
x
The fixed-fixed case gives
Z 1
Z x
x2 1
x2
1
(1−x)ada+ (1−a)xda = (1−x) + x−(x− )x = (x−x2 ).
u(x) =
2 2
2
2
x
0
(3)
Solution 2. The quadratic spline is obtained as the primitive of the ramp
(
0,
x≤0
Q(x) = 1 2
(4)
x≥0
2x ,
It has one continuous derivative at zero (the ramp is continuous at zero).
The cubic spline is obtained as the primitive of the quadratic spline
(
0,
x≤0
Q(x) = 1 3
(5)
x
,
x
≥0
6
It has two continuous derivative at zero.
Solution 3. By integration, the general form of u(x) is
u(x) = C(x) +
A 3 B 2
x + x + Cx + D.
6
2
Date: February 29, 2016.
1
(6)
2
SOLUTION SET (PSET 2, 18.085)
Now we use the four boundary conditions
A B
+ + Cx + D,
6
2
A B
0 = u(−1) = C(−1) − + − Cx + D,
6
2
0 = u00 (1) = R(1) + A + B,
0 = u(1) = C(1) +
(7)
0 = u00 (−1) = R(−1) − A + B,
Note that C(1) = 16 , C(−1) = 0, R(1) = 1, R(−1) = 0. We can substitute
that into the equations above and solve for A, B, C, D. The result is
u(x) = C(x) −
1 3 1 2 1
x − x + .
12
4
6
(8)
Solution 4. Our 4-by-3 backward difference matrix is


−1 0
0
 1 −1 1 

∆− = 
0
1 −1
0
0
1
By direct computation we can verify that




−1 0
0 
1 −1 0
0 0
 1 −1 1  −1 1
−1 2
1
  0 −1 1 0 = 
B4 = 
0


1 −1
0 −1 2
0
0 −1 1
0
0 −1
0
0
1
(9)

0
0
 , (10)
1
1
and



 −1 0


0
−1 1
0 0 
2
−1
0
1 −1 1 
 = −1 2 −1 .
K3 =  0 −1 1 0 
0
1 −1
0
0 −1 1
0 −1 2
0
0
1
A straightforward MATLAB computation given eigenvalues
• for K3 : 0.5858,2,3.4142
• for B4 : 0,0.5858,2,3.4142
which are the squared singular values of ∆− .
(11)
Solution 5. Because determinant is multiplicative, given such a decomposition, we have
det(A) = det(SΛS −1 )
= det(S) det(Λ) det(S −1 )
= det(S) det(Λ) det(S)−1
= det(Λ) = λ1 · . . . · λn ,
(12)
SOLUTION SET (PSET 2, 18.085)
3
where Λ = diag(λ1 . . . λn ). By definition, such a decomposition exists iff A
is diagonalizable.
Solution 6. C = At A is only semi-definite. We see it easily because A is
singular, hence there exists a nonzero vector in the null space, i.e., ∃u such
that Au = 0, hence Cu = 0 as well. To find all such vectors, we solve

   
1 −1 0
u1
0
0




1 −1
u2 = 0 .
(13)
−1 0
1
u3
0
But that’s just u1 = u2 = u3 . So
 
c
null(A) = c , c is any constant.
c
(14)
Solution 7. We use the determinant test for positive definite and then
pivot and multiplicities:
•
1 b
A=
⇒ 1 > 0, 9 − b2 > 0,
(15)
b 9
So the condition is −3 < b < 3. The pivots are
1
b
U=
(16)
0 9 − b2
The multiplicities
1 0
L=
b 1
Indeed, the factorization is
1
b
1 b
1 0
A=
b 1
0 1
0 9 − b2
(17)
(18)
•
2 4
A=
⇒ 2 > 0, 2c − 16 > 0,
4 c
So the condition is c > 8. The pivots are
2
4
U=
0 c−8
(19)
(20)
The multiplicities
1 0
L=
2 1
Indeed, the factorization is
1 2
2
4
1 0
A=
0 1
0 c−8
2 1
(21)
(22)
4
SOLUTION SET (PSET 2, 18.085)
Solution 8. First proof: if the eigenvalues of A are λ1 , . . . , λn , then the
−1
eigenvalues of A−1 are λ−1
1 , . . . , λn . But if all the eigenvalues of A are
positive, then so are their reciprocals.
Second proof: we use the determinant test of the entries of the inverse matrix,
which is given by the formula
1
a −b
−1
(23)
A =
ac − b2 −b c
The first determinant we must check is the upper left corner. So for A−1 to
be positive definite, we need to have
c
> 0.
(24)
ac − b2
Then we also need
ca − (−b)2
> 0.
(25)
ac − b2
But that holds automatically if A is positive definite! For, if A is positive
definite then ac − b2 = det(A) > 0. The pivot c − ab b is positive as well,
2
2
which implies c > ba > 0. The next equation is actually 1 = ca−b
> 0 which
ac−b2
holds.
Solution 9. Consider the product

 
 
a1
4 1 1
a1
a1 a2 a3 1 0 2 a2  = 4a1 + a2 + a3 , a1 + 2a3 , a1 + 2a2 + 5a3 a2 
1 2 5
a3
a3
= 4a21 + 2a1 a2 + 2a1 a3 + 4a2 a3 + 5a23
(26)
For example, we can take a1 = 0, a2 = 1 and a3 = 0. We get zero as a result
which is impossible.
Solution 10. The quadratic form in the question is just
a b
x
z = f (x, y) = x y
= ax2 + 2bxy + cy 2 .
b c
y
(27)
To have a saddle point, we need one positive eigenvalue and one negative.
Equivalently the determinant of the 2-by-2 matrix, which is the product of
the eigenvalues, must be negative. So the test is ac − b2 < 0.
Solution 11. M is a block triangular matrix so it’s determinant is the
product det(H) · deg(K) which is non-zero. N must have linearly dependent
columns so it is singular and can not be positive definite. In another way,
det(N ) = det(K − KK −1 K) = 0. The pivots of M are those of H and K.
The pivots on N are the pivots of K. Similarly, the eigenvalues of M are
the eigenvalues of K and H, while the eigenvalues of M are the eigenvalues
SOLUTION SET (PSET 2, 18.085)
5
of K, just with double the algebraic multiplicity. The Cholesky factorization
of M is
chol(H)
0
chol(M ) =
.
(28)
0
chol(K)
because this is upper triangular and by definition
chol(H)
0
chol(H)
0
H 0
=
(29)
0
chol(K)
0
chol(K)
0 K
but that’s just M !