18.085 PROBLEM SET 2 SOLUTIONS
HAOFEI WEI
Problem (1.6.1).
u
Solution. Let u = 1 .
u2
T
u Tu =
u1
=
u1
1 −1 u1
u2
−1 2
u2
u1 − u2
u2
−u1 + 2u2
= u21 − 2u1 u2 + 2u22
= u22 + (u1 − u2 )2 ≥ 0
The product equals zero only when u = 0, which shows that T is positive definite.
Problem (1.6.9).
Solution. Since we need to factor the matrices into LDLT form anyways, we will factor them
and use the pivots to solve for b and c.
1 b
1
b
Using simple row reduction, we can reduce
to
. Then, by inspection,
b 9
0 9 − b2
1 0
1 b
1 0 1
0
1 b
we can find that L =
. Thus, we have
=
. For this
b 1
b 9
b 1 0 9 − b2 0 1
matrix to be positive definite, we need all pivots to be positive, and thus the necessary
condition is −3 < b < 3. 2 4
1 0 2
0
1 2
Similarly, we can factor
into
. Thus, the condition for the
4 c
2 1 0 c−8 0 1
matrix to be positive definite is c > 8.
Problem (1.1.27).
Solution. For the first matrix and second matrices, the two columns are not multiples of each
other (column1 6= c × column2 ). Thus, the matrix AT A is positive definite.
For the third matrix, we have two rows and three columns, so they must be linearly
dependent
(3 vectors
in 2-dimensional space). We can also directly show this by seeing that
1
2
1
+
=3
. Thus, the matrix formed by AT A is not positive definite.
2
1
1
Problem (1.6.20).
1
HAOFEI WEI
18.085 PROBLEM SET 2 SOLUTIONS
cos(θ) − sin(θ)
2 0
Solution. For simplicity’s sake, we will let Q =
and Λ =
.
sin(θ) cos(θ)
0 5
− sin(θ)
2
2
Note first that |Q| = cos (θ) + sin (θ) = 1. Next, we see that cos(θ) sin(θ)
=
cos(θ)
0. Thus, M is an orthogonal matrix, and so A must be expressed in its eigenvalue decomposition form.
(a) det(A) = det(Q) det(Λ) det(QT ) = (1)(10)(1) = 10.
(b) Λ is the eigenvalue matrix of A, so the eigenvalues of A are 2 and 5.
cos(θ)
(c) The eigenvectors of A are the columns of Q, so the two eigenvectors are
and
sin(θ)
− sin(θ)
, corresponding to the eigenvalues 2 and 5, respectively.
cos(θ)
(d) AT = (QT )T ΛT QT = QΛQT = A. Since A is symmetric, and its eigenvalues are
positive, A must be positive definite.
Problem (1.6.27).
Let m, n such that H is an m × m matrix and K is an n × n matrix. Next, let λH,i be
the set of m positive eigenvalues of H, not necessarily unique, and λK,i be similarly defined
for K. Then, let vH,i be the eigenvector of H corresponding to λH,i , and vK,i be similarly
defined for K.


λH,1


λH,2
, such that
Finally, let QH = vH,1 vH,2 . . . vH,m and ΛH = 
..


.
λH, m
QH ΛH QTH .
Let
QK and ΛK be similarly
definedfor
TK.
H
QH
ΛH
QH
We can factor
into
≡ QΛQT . By inspection,
K
QK
ΛK
QTK
we can see that Q is orthogonal and Λ is a diagonal matrix, so QΛQT factors M into its
eigenvalue decomposition. Since ΛH and ΛK both have only positive diagonal entries, Λ
also has all positive diagonal entries. Thus, M is positive definite, and its eigenvalues are
{λH,i } ∪ {λK,i }.
To find the eigenvalues of N isslightly more involved. By inspection and through some
v
trial and error, we see that K,i is an eigenvector of N with eigenvalue 2λK,i . Similarly,
vK,i
vK,i
is an eigenvector of N with eigenvalue 0. Thus, we have found 2n eigenvectors for
−vK,i
N , which is a 2n × 2n matrix, and so we have found all its eigenvalues - {2λK,i } ∪ {0}.
Next, we must find the pivots of the two matrices. The structure of M is such that row
operations of one sub-matrix do not affect the rows of the other sub-matrix.
Thus, if we
UH 0
perform elementary row operations on M , we will reduce M to
, where UH and
0 UK
UK are the reduced forms of H and K with the pivots on the diagonal. Thus, the pivots
of
M are the union of sets of pivots of H and K. Similar operations on N will reduce it to
UH UH
, which shows that the pivots of N are the pivots of K, along with n zeroes.
0
0
2
H=
HAOFEI WEI
18.085 PROBLEM SET 2 SOLUTIONS
Our last task is to find
define
T chol(M ). We will
chol(H)
= A and chol(K) = B. Thus,
A A
AT 0
A 0
we can factor M into
=
. Thus, by inspection, we can see
BT B
0 BT 0 B
chol(H)
0
that chol(M ) =
.
0
chol(K)
Solution.
Problem (2.1.5).

1
0
0
−1 1
0

Solution. For a fixed-fixed spring-mass system, we know AT CAu = f , where A = 
 0 −1 1 ,
0
0 −1

 

u1
c1




c2
, u is the displacement vector u2 , and f is the external force vecC =

u3 

c3
c4
u4
 
mg
tor mg . To simplify the algebra, which would otherwise be extremely messy, we will
mg
set c1 = c2 = c3 = c4 = c. Then,
using MatLab’s “\” command, we can solve the sys 
3
4. From that, we can find the forces on the springs
tem to find u = AT CA\f = mg
2c
3
 
3


 1 . Be very careful about what the meaning of the signs are. A positive
W = CAu = mg
2c −1
−3
displacement means displacement downwards, and a positive force means a force pushing
outwards from the center of the spring.
The reaction force from the ceiling and floor will be equal in magnitude and opposite in
direction to the forces from the first and fourth springs. Thus, the reaction force from the
ceiling will be 3mg
upwards to counteract the downward pulling of spring 1, and the reaction
2
force from the floor will 3mg
upwards to counteract the downward pushing from spring 4.
2
Thus, the total reactionary force will be 3mg upwards, which perfectly opposes the 3mg
downwards from gravity acting on the 3 masses.

Problem (2.1.7).


1 0
0
Solution. For c1 = c3 = c4 = 1 and c2 = 0, we have K = AT CA = 0 1 −1. The
0 −1 1
determinant of this matrix is 1 (easily 
found
with
MatLab),
so
it
is
invertible.

 
1
1



We can solve the system Ku = f = 1 to find u = K\f = 3.
1
2
3
HAOFEI WEI
18.085 PROBLEM SET 2 SOLUTIONS
To analyse
  this system in more detail, we can look at the forces on this system. w =
1
0

CAu = 
−1. c2 being 0 means that the system is behaving as if the second spring is not
−2
there, and the system is decoupled between the first and second masses. The first spring’s
force is 1, indicating that it’s only supporting the top mass. Spring 3’s force is also 1, which
makes sense because it’s also only supporting one mass - mass two. The force in spring 4 is
2, as it is supporting only masses two and three.
4