Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
The following problems are for additional practice and are not to be turned in: (All
problems come from Linear Algebra: A Geometric Approach, 2nd Edition by ShifrinAdams.)
Exercises: Section 2.1: 1–5, 7, 8, 14
Section 2.2: 1–7, 10, 11
Turn in the following problems.
1. Divide the interval [0, 1] into five equal subintervals, and apply the finite difference method in order to approximate the solution of the two-point boundary value
problem
y 00 (t) = 125t,
y(0) = y(1) = 0,
at the four interior grid points. Compare your approximate values at the grid points
with the exact solution at the grid points. Note: You should not expect very
accurate approximations with only four interior grid points. The exact solution is
125 3
(t − t).
6
Solution: Following the discussion from class, we start by subdividing the
interval [0, 1] into five subintervals. Note that h = 1−0
= 51 Using the naming
5
convention that we introduced in class, we have t0 = 0, y0 = y(t0 ), t1 =
1/5, y1 = y(t1 ), t2 = 2/5, y2 = y(t2 ), t3 = 3/5, y3 = y(t3 ), t4 = 4/5, y4 = y(t4 )
and t5 = 1, y5 = y(t5 ). Notice that y0 = y(0) = 0 and y5 = y(1) = 0 from our
boundary conditions and thus our unknowns are y1 , y2 , y3 and y4 . Using the
following approximation from class (Note that h = 1/5)
y 00 (t) ≈
y(t + h) − 2y(t) + y(t − h)
= 125t =⇒
h2
y(t + h) − 2y(t) + y(t − h) = h2 ∗ 125t = 5t
we obtain the following system of equations:
y2 − 2y1 + y0
y3 − 2y2 + y1
y4 − 2y3 + y2
y5 − 2y4 + y3
= 5(1/5)
= 5(2/5)
= 5(3/5)
= 5(4/5)
,
1
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
which in matrix form (recall that y0 = y5 = 0) is given by:

   
−2 1
0
0
y1
1
 1 −2 1
 y2   2 
0

  =  
0
1 −2 1  y3   3 
0
0
1 −2 y4
4.
To find the solution, we can perform Gaussian elimination to obtain:

−2 1
0
0
 1 −2 1
0
[A|b] = 
 0
1 −2 1
0
0
1 −2


1
1 0
 0 1
2 
→
 0 0
3 
4
0 0
0
0
1
0
0
0
0
1

−4
−7 

−8 
−6
Thus we have that
y1 = −4
y2 = −7
y3 = −8
y4 = −6.
Comparing to the exact solution, we find that y(1/5) = −4, y(2/5) = −7, y(3/5) =
−8, y(4/5) = −6.
2. Divide [0, 1] into n + 1 equal subintervals and apply the finite difference approximation method to derive the linear system associated with the two-point boundary
value problem
y 00 (t) − y 0 (t) = f (t),
y(0) = y(1) = 0.
We proceed as in class and begin by first subdividing the interval [0, 1] into n + 1
b−a
equal subintervals so that h = n+1
. To convert the differential equation to a
linear system of equations, we use the following approximations for y 00 (t) and
y 0 (t):
2
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
y(t + h) − 2y(t) + y(t − h)
h2
y(t + h) − y(t)
y(t + h) − y(t − h)
or
y 0 (t) ≈
2h
h
y 00 (t) ≈
Using the naming convention, yi = y(ti ) and fi = f (ti ) and the fact that ti =
ti−1 + h we obtain
yi+1 − 2yi + yi−1
h2
yi+1 − yi−1
yi+1 − yi
y 0 (ti ) ≈
or
,
2h
h
y 00 (ti ) ≈
which yields the following system of equations (by replacing y 00 (t) − y 0 (t) = f (t)
with its approximation) with unknowns given by y1 , y2 , . . . , yn (y0 = yn+1 = 0):
y0 − 2y1 + y2 − h2 (y2 − y0 )
y1 − 2y2 + y3 − h2 (y3 − y1 )
h
= −2y1 + (1 − )y2 = h2 f1
2
h
h
= (1 − )y1 − y2 + (1 − )y1 = h2 f2
2
2
..
.
h
h
= (1 − )yn−2 − yn−1 + (1 − )yn = h2 fn−1
2
2
h
yn−1 − 2yn + yn+1 − h2 (yn−1 − yn+1 ) = (1 − )yn − 1 − yn = h2 fn .
2
yn−2 − 2yn−1 + yn − h2 (yn−2 − yn )
This yields the following matrix equation:
  y   h2 f 
1
1
−2 1 − h2
0
0
0
...
0
2




h
h
y
h
f
1 − 2 −2 1 − 2
2 
2 
0
0
...
0 



.
.




h
h
 0
1 − 2 −2 1 − 2
0
...
0 
 ..   .. 
 .



..
..
..  
..
..
..
 .
.. 
 =  ... 
.
.
.
.
.
. 
 .
.

 

 .
 . 
..
..
.. 
...
...
..
.. 
 ..

.
.



.
.
.
.   . 



h 
h
 0
0
0
. . . 1 − 2 −2 1 − 2 yn−1  h2 fn−1 
0
0
0
0
. . . 1 − h2 −2
yn
h2 fn

3. Section 2.1, Problem 6
3
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
We wish to prove or give a suitable counterexample for the following claims for n×n
matrices:
(a) If AB = CB and B 6= 0, then A = C.
1 2
0
False. Consider the matrices A =
,B=
2 3
1
1 2 2 2
2 0
Then AC =
=
and CB =
2 3 2 3
3 0
but A 6= C.
0
0
2
2
2 2
and C =
.
2 3
2 0 0
2 0
=
3 1 0
3 0
(b) If A2 = A, then A = 0 or A = I.
0 0
0 0 0 0
0 0
2
False. Let A =
. Then A = AA =
=
= A but
0 1
0 1 0 1
0 1
A 6= 0 and A 6= I.
(c) (A + B)(A − B) = A2 − B 2
False. In order for this statement to be true, we would need AB = BA,
since (A + B)(A − B) = A2 − AB + BA − B 2 . However, in general matrix
multiplication
is not
To see this,consider thematrices,
communative.
A=
1 2
−1 0
1 0
−1 −2
and B =
. Then AB =
but BA =
.
−3 1
1 0
4 0
1
2
(d) If AB = CB and B is nonsingular, then A = C.
True. To prove this, we suppose that AC = BC and that B is nonsingular.
Since B is nonsingular, we know that for any b ∈ Rn there is a unique
x ∈ Rn such that Bx = b. Since this is true for any b ∈ 
Rn
, if I let
1
0
 
 
b1 = (1, 0, 0, . . . , 0), then we know there is an x1 so that Bb = 0. Using
 .. 
.
0
the fact that AB = CB we have that
 
 
1
1
0
0
 
 
 
 
AB = CB =⇒ ABx1 = CBx1 =⇒ A 0 = C 0 .
 .. 
 .. 
.
.
0
0
4
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
 
1
0
 
 
By definition of matrix multiplication, A 0 = first column of A, and
 .. 
.
0
 
1
0
 
 
C 0 = first column of C. So the first column of C is precisely the first
 .. 
.
0
column of A. To show that the ith column of A is equal to the ith column
of C, we choose bi = (0, 0, . . . , 1, 0, . . . 0) where 1 is the ith position of bi .
Then there is an xi such that Bxi = bi . Using the same procedure as above,
we can show that the ith column of A is equal to the ith column of C. Since
every column in A is the same as every column in C, it must be the case
that A = C.
(e) If AB = BC and B is nonsingular, then A = C.
1 2
This claim is false. Consider the following matrices: A =
, B =
3 4
1 0
1 −2
and C =
. Note that B is nonsingular, as it has rank 2
0 −1
−3 4
(it is in echelon form and has 2 nonzero rows). We also see that AB = BC,
but A 6= C.
4. Section 2.1, Problem 10
We wish to show that if A and B are nonsingular n×n matrices, then the product
AB is nonsingular.
Proof : We show that AB is nonsingular by showing that ABx = 0 =⇒ x = 0.
Suppose that ABx = 0. Then ABx = A(Bx) = 0 =⇒ Bx = 0 since A
is nonsingular. Now B is also nonsingular, so Bx = 0 =⇒ x = 0. Thus,
ABx = 0 =⇒ x = 0 and so AB is nonsingular.
5. Section 2.1, Problem 11
We prove the following statements:
(a) If A is an m × n matrix and B is an n × m matrix such that BA = In , then if
for some b ∈ Rm , Ax = b has a solution, then that solution is unique.
5
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
Proof : To prove this, let us suppose that there are two such solutions to
Ax = b, call them x1 and x2 . Then Ax1 = b and Ax2 = b. Multiplying
both expressions by B, we obtain BAx1 = Bb =⇒ In x2 = Bb =⇒ x1 =
Bb. Similarly, BAx2 = Bb =⇒ In x2 = Bb =⇒ x2 = Bb. So x1 = x2
and hence the solution is unique.
(b) If A is an m × n matrix, are C is an n × m matrix such that AC = Im , then
Ax = b is consistent for every b ∈ Rm .
Proof: We claim that for every b ∈ Rm , x = Cb is a solution to Ax = b.
We verify that this is indeed the case as follows: Ax = A(Cb) = (AC)b =
Im b = b.
(c) If A is an m × n matrix and B and C are n × m matrices such that BA = In
and AC = Im , then B = C.
Proof: Suppose that A is an m × n matrix and B and C are n × m matrices
such that BA = In and AC = Im . Then B = BIm = B(AC) = (BA)C =
In C = C.
6. (a) Suppose that A and B are two n × n upper triangular matrices. Prove that
the product AB is upper triangular.
Proof : Suppose that A and B are n × n upper triangular matrices. Then
all entries of A and B below the diagonal are 0. That is, aij = 0 = bij if
i > j. We we need to show is that the (i, j) entry of AB is equal to 0 as
long as i > j. By definition,
(i, j) entry of AB = (row i of A) · (column j of B)
= [ai1 , ai2 , . . . , ain ] · [b1j , b2j , . . . , bnj ]
= ai1 b1j + ai2 b2j + . . . + ain bnj .
Assume that i > j. We show that the (i, j) entry of AB is 0. It would clearly
suffice to show that either aik or bkj is 0 for all k. Consider two possibilities.
k > j and k ≤ j. If k > j, then bkj represents an entry below the main
diagonal of B. Therefore, bkj = 0 and so the summand aik bkj = 0. Now if
k ≤ j, then since j < i, i > k. Therefore, aik represents an entry below the
main diagonal of A, and aik = 0. Thus, aik bkj = 0. Since aik bkj = 0 for all
k, we see that
(i, j) entry of AB = ai1 b1j + ai2 b2j + . . . + ain bnj = 0,
6
Math 317: Linear Algebra
Homework 3 Solutions
Fall 2015
and so AB is upper triangular as required.
(b) If A and B are two n × n upper triangular matrices, what are the diagonal
entries of AB?
When i = j, the only nonzero term in the product (row i of A)·(column j of B)
is aii bii .
(c) Is it true that the product of two n × n lower triangular matrices is again lower
triangular? Justify your answer.
Yes, you can just repeat the proof of (a) using the fact that i < j now
instead of i > j.
7