Math 317: Linear Algebra
Homework 4
Due: September 30th, 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 2.2: 1–7, 10, 11
Section 2.3: 1–4, 6–8
Turn in the following problems.
1. Section 2.2, Problem 3b,3c
Solution, 2.2, 3b: To find
the standard matrix A for the linear transformation
1
0
T , we need to compute T
(which is the first column of A) and T
0
1
(the second column of A). We are given a linear transformation T with the
values:
2
2
0
1
T
=
T
=
.
1
−3
1
−3
1
To calculate T
, we recall that for any linear transformation T : R2 →
0
R2 we have that
1
0
x1
= x1 T
+ x2 T
.
T
x2
0
1
Thus, we have that
5
2
1
1
=T
= 2T
+1
.
3
1
0
−3
1
2
Solving this vector equation yields: T
=
. Thus the standard
0
3
matrix A for this linear transformation is given by:
2 1
A=
.
3 −3
Solution, 2.2, 3c: This time we are given the following values for a linear
transformation T :
1
3
T
=
1
3
T
1
−1
=
.
−1
1
1
Math 317: Linear Algebra
Homework 4
Due: September 30th, 2015
1
0
To find T
and T
, we once again use the fact that for any linear
0
1
transformation T : R2 → R2 we have that
x1
1
0
T
= x1 T
+ x2 T
.
x2
0
1
This, yields the following system of equations:
1
3
1
0
T
=
=T
+T
1
3
0
1
1
−1
1
0
T
=
=T
−T
.
−1
1
0
1
Solving this system of equations yields the following:
1
1
0
2
T
=
, and T
=
.
0
2
1
1
Thus the standard matrix A for this linear transformation is given by:
1 2
A=
.
2 1
2. Section 2.2, Problem 5a,5c,5d,5f
Solution, 2.2, 5a We wish to give the 2 × 2 matrix A so that for any x ∈ R2 we
have that Ax is the vector whose components are, respectively, the sum and
difference of the components of x. That is we want to find a matrix such that:
a b x1
x1 + x2
=
c d x2
x1 − x2
By choosing a = 1, b = 1, c = 1, d = −1, we
see that
the matrix multiplication
1 1
gives us exactly what we wanted, so A =
.
1 −1
Solution, 2.2, 5c We wish to give the 2 × 2 matrix A so that for any x ∈ R2 we
have that Ax is the vector obtained by first reflecting x across the line x1 = 0
and then reflecting the resulting vector across the line x2 = x1 . Because we
have two different transformation, we need to find a matrix A1 such that A1 x
is the result of reflecting x across the line x1 = 0, and then a matrix A2 so
that A2 (A1 x) is the result of reflecting A1 x across the line x2 = x1 . We see
that the final answer will be A = A2 A1 . Reflecting across the line x1 = 0 is
2
Math 317: Linear Algebra
Homework 4
Due: September 30th, 2015
equivalent to reflecting about the y-axis. Thus, we are looking for a matrix A1
such that
a b x1
−x1
A1 x =
=
.
c d x2
x2
−1 0
We see that A1 =
does the job for us. You could have also used the
0 1
formula Rl (x) = 2Pl (x) − x and found the standard matrix associated with
this linear transformation. Now, to find A2 , we look for a matrix such that
a b −x1
a b
x1
x2
A2 (A1 x) =
=
A
=
c d
x2
c d 1 x2
−x1
0 1
We see that A2 =
will do the job for us. Thus A = A2 A1 =
1 0
0 1 −1 0
0 1
=
.
1 0
0 1
−1 0
Solution, 2.2, 5d This is precisely problem 3c on Exam 1. Please refer to the
exam solutions for how to do this problem.
Solution, 2.2, 5f We wish to give the 2 × 2 matrix A so that for any x ∈
R2 we have that Ax is the vector obtained by first rotating x an angle of
π/2 counterclockwise and then projecting the resulting vector onto the line
x1 + x2 = 0. Similar to 2.2, 5c, the final answer A will be written as a product
A = A2 A1 , where A1 is the standard matrix associated with the rotation, and
A2 is the standard matrix associated with the projection. In class, we defined
the rotation matrix as
− sin θ
,
cos θ
0 −1
. A2 is the same matrix in
1 0
1/5 2/5
3c) which is given as A2 =
. Thus A =
2/5
4/5
−1
2/5 −1/5
=
.
0
4/5 −2/5
cos θ
A=
sin θ
cos π/2 − sin π/2
and thus A1 =
=
sin π/2 cos π/2
Problem 5d (Exam 1,
1/5 2/5 0
A2 A1 =
2/5 4/5 1
3. Section 2.2, Problem 9
Proof : Recalling that Rl (x) = xk − x⊥ , we want to show that kRl (x)k = kxk.
To this extent, we square both sides, and use the relationship between norms
and dot product to write:
3
Math 317: Linear Algebra
Homework 4
Due: September 30th, 2015
kRl (x)k2 = kxk − x⊥ k2
= x k − x⊥ · xk − x⊥
= (xk · xk ) − 2(xk · x⊥ ) + (x⊥ · x⊥ )
= (xk · xk ) + (x⊥ · x⊥ ),
since x⊥ is perpendicular to a, and xk is parallel to a (so xk ·x⊥ = 0). However,
we also have that
kxk2 = kxk + x⊥ k2
= xk + x⊥ · xk + x ⊥
= (xk · xk ) + 2(xk · x⊥ ) + (x⊥ · x⊥ )
= (xk · xk ) + (x⊥ · x⊥ ).
Thus kRl (x)k = kxk.
To show that Rl (x) · a = x · a, we observe the following:
Rl (x) · a =
=
=
xk − x ⊥ · a
xk · a − x⊥ · a
xk · a .
Similarly, we have that
x·a =
=
=
x k + x⊥ · a
xk · a + x⊥ · a
xk · a .
Thus Rl (x) · a = x · a.
4. Section 2.2, Problem 10
Proof : We observe that T (0) = T (0 + 0) = T (0) + T (0) since T is a linear
transformation. So T (0) = 2T (0) =⇒ T (0) = 0.
To prove that T (au+bv) = aT (u)+bT (v) for all u, v ∈ Rn and scalars a, b ∈ R,
4
Math 317: Linear Algebra
Homework 4
Due: September 30th, 2015
we can use the properties of linear transformations to say that T (au + bv) =
T (au) + T (bv) = aT (u) + bT (v).
5. Section 2.3, Problem 2c
We find the inverse of A using Gaussian elimination as follows:



1 1 1 1 0 0
1
−R1 +R3 →R3



0
[A|I] = 0 1 1 0 1 0
−→
1 2 1 0 0 1
0



1 0 0
1 −1 0
1 0
−R3 →R3
 0 1 1


0
1 0
0 1
−→
R3 +R2 →R2
0 0 −1 −1 −1 1
0 0


1 −1 0
1 .
so, A−1 = −1 0
1
1 −1

1 1 1 0 0
−R2 +R3 →R3
1 1 0 1 0 
−→
−R2 +R1 →R1
1 0 −1 0 1

0 1 −1 0
1  = [I|A−1 ],
0 −1 0
1 1
1 −1
Once we are equipped with the inverse, we know that we can solve Ax = b by
taking
both sides to obtain x = A−1 b. Thus, the solution
 the inverse
  of A on
 
1 1 1
x1
3





x2 = 0 is
to 0 1 1
1 2 1
x3
1

   
1 −1 0
3
3
1  0 = −2 .
x = A−1 b = −1 0
1
1 −1 1
2
This allows us to write b as a linear combination of the columns of A as follows:
 
 
 
1
1
1
b = 3 0 − 2 1 + 2 1 .
1
2
1
6. Section 2.3, Problem 9
Proof : To prove that A and A − I are invertible, we find matrices B1 and
B2 such that AB1 = I and (A − I)B2 = I, given that A satisfies the matrix
equation: A3 − 2I = 0. To show that A is invertible, we have that:
1
A − 2I = 0 =⇒ A = 2I =⇒ A3 = I =⇒ A
2
3
3
1 2
A
2
= I.
So B1 = 12 A2 and hence A is invertible. To show that A − I is invertible, we
have that:
5
Math 317: Linear Algebra
Homework 4
Due: September 30th, 2015
A3 − 2I = 0 =⇒ A3 − I = I =⇒ (A − I)(A2 + A + I) = I,
and so B2 = A2 + A + I. Thus A − I is invertible.
7. Section 2.3, Problem 11
Proof : We prove that if AB is nonsingular then A is nonsingular, and B is
nonsingular. We begin by showing that B is nonsingular. Suppose that Bx =
0. We show that the only solution to this system is the trivial solution. That
is, x = 0. Now Bx = 0 =⇒ A(Bx) = A(0) = 0 =⇒ (AB)x = 0 =⇒ x = 0
since we know by assumption that AB is nonsingular. So Bx = 0 =⇒ x = 0
and hence B is nonsingular. To show that A is nonsingular, we prove that A
can be written as the product of two nonsingular matrices. We recall that the
product of nonsingular matrices is nonsingular. Now since B is nonsingular,
then B is invertible and B −1 exist. Furthermore, B −1 is invertible (and hence
nonsingular). Using the fact that B −1 is nonsingular and AB is nonsingular
by assumption, we write A as A = (AB)B −1 . Since A is the product of two
nonsingular matrices, then A itself must be nonsingular.
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