MATH 260 –Homework 1 solutions
1. At noon the minute and hour hands of a clock coincide.
(a) What is the first time, T1 , when they are perpendicular?
(b) What is the next time, T2 when they again coincide?
Let time t be measured in hours. Then the the hour hand travels 2π radians in 12 hours, so the
angle it makes with the vertical (12 o’clock position) is θH ptq πt{6. Likewise, the minute hand
travels 2π radians in one hour, so θM ptq 2πt.
(a) We want to know when θM ptq θH ptq π {2, in other words, when p2π π {6qt π {2. This
happens when 11t{6 1{2, or t 3{11. This means 3{11 hours past noon, or 180{11 minutes –
which is 16 minutes, 21.8181. . . seconds past noon.
(b) We could do this the same way as part (a) (except we would want to know when θM ptq θH ptq 2π)), but another way is to realize that the hands come together sometime between 1 and 2
o’clock, then again sometime between 2 and 3 o’clock and so forth up until sometime between 10 and
11 o’clock. So the times the hands come together divide the 12 hours between noon and midnight
into eleven equal intervals. Therefore, the first time the hands come together is 12/11 hours past
noon, or 1 hour, 5 minutes, 27.2727. . . seconds past noon.
4. For which real numbers x do the vectors px, 1, 1, 1q, p1, x, 1, 1q, p1, 1, x, 1q, p1, 1, 1, xq not form a
basis of 4 ? For each of the values of x that you find, what is the dimension of the subspace of 4
that they span?
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The vectors form a basis of 4 if and only if they are linearly independent (since there are four
of them). So we need to determine whether the equation
apx, 1, 1, 1q
bp1, x, 1, 1q
has any solutions besides a b c d
backwards to make calculations easier):
d
d
d
xd
cp1, 1, x, 1q
0.
c
c
xc
c
dp1, 1, 1, xq p0, 0, 0, 0q
This gives us four equations to solve (we write them
b
xb
b
b
xa
a
a
a
0
0
0
0
Subtract the first equation from each of the next two, and x times the first equation from the last
to obtain:
d
c
b
xa 0
px 1qb
p1 xqa 0
px 1qc
p1 xqa 0
p1 xqc
p1 xqb
p 1 x2 q a 0
2
Next, move the second equation to the bottom and then add the new second equation to the new
third equation to obtain:
d
c
xa 0
p 1 xq a 0
p 2 x x2 q a 0
p1 xqa 0
b
px 1qc
p1 xqb
px 1qb
Next add the third equation to the fourth equation to get
d
c
b
px 1qc
p1 xqb
xa 0
p 1 xq a 0
p 2 x x2 q a 0
p3 2x x2 qa 0
Since 3 2x x2 p3 xqp1 xq, these equations have only the zero solution unless x is either
1 or 3. If x 1, then the last three equations reduce to 0 0, and so a, b and c can be chosen
arbitrarily and then d is determined — we can also see that the four vectors all become identical
(equal to p1, 1, 1, 1q), so the dimension of the subspace is 1.
When x 3, then once we choose d then c, b and a are determined — therefore the dimension
of the subspace spanned by the vectors p3, 1, 1, 1q, p1, 3, 1, 1q, p1, 1, 3, 1q and p1, 1, 1, 3q is three
(it is the set of vectors pa, b, c, dq with a b c d 0).
We also note that the determinant of the matrix is px 1q3 px 3q — which is zero when x 1
or x 3. It is not a coincidence that x 1 is a triple zero of the determinant and the dimension
of the corresponding subspace is 4 3 1 and x 3 is a simple zero of the determinant and the
dimension of the corresponding subspace is 4 1 3.
6. Compute the dimension and find bases for the following linear spaces.
(a) Real anti-symmetric 4 4 matrices.
(b) Quartic polynomials with the property that pp2q 0 and pp3q 0.
(c) Cubic polynomials ppx, y q in two real variables with the properties: pp0, 0q
and pp0, 1q 0.
(d) The space of linear maps L :
R5 Ñ R3 whose kernels contain p0, 2, 3, 0, 1q.
(a) A real anti-symmetric 4 4 matrix looks like:
0
a
a
0
b d
c
b c
d e
0 f
e f 0
0, pp1, 0q 0
3
which clearly has six independent parameters, so the dimension is six. A basis is
0
1
0
1 0 0
0 0 0
0 0 0
0 0 0 0
0 0
0 0
,
1 0
0 0
0
0
0
0
0 1
1 0
0 0
0 0
0 0
0 0
1 0
0 0
0 0 0
0
0
0
,
0
1
0
0
,
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0 0 0
0 0 1
0 0 0
1 0 0
,
,
0
0
0
0
0
0
0
0
0 0
0 0
0 1
1 0
.
(b) A polynomial ppxq satisfies pp2q 0 and pp3q 0 if it has px 2q and px 3q as factors. So
a quartic (fourth-degree) polynomial satisfies this if and only if it is the product of px 2qpx 3q
with a quadratic polynomial a0 a1 x a2 x2 . Therefore, the dimension of this space is three, and
it has basis
px 2qpx 3q, xpx 2qpx 3q, x2 px 2qpx 3q.
(c) A cubic polynomial in two variables looks like:
ppx, y q a0
a1 x
a3 x2
a2 y
a5 y 2
a4 xy
a6 x3
a7 x2 y
a8 xy 2
a9 y 3 .
Since pp0, 0q a0 , pp1, 0q a0 a1 a3 a6 and pp0, 1q a0 a2 a5 a9 and these are independent
equations, the dimension of the space of cubic polynomials satisfying pp0, 0q pp1, 0q pp0, 1q 0
is 10 3 7. A basis is:
x x2 ,
y y2 ,
xy,
(d) The space of maps from
x2 x3 ,
x2 y,
R5 to R3 has dimension 15:
a11
a21
a31
a12
a22
a32
a13
a23
a33
a14
a24
a34
a15
a25
a35
xy 2 ,
y2 y3 .
.
If p0, 2, 3, 0, 1q is in the kernel, then we have the equations
2a12 3a13
0, 2a22 3a23 a25 0,
so the dimension of the space is 15 3 12 and a basis is
1 0 0 0 0
0 1 0 0 2
0 0 1 0
0
0
0
0
0
0
0
0
0
0
,
a15
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
2a32 3a33
,
0 , 0 0 0 0
0
0 0 0 0
0
0 0 0 0 0
0 , 0 1 0 0 2 , 0
0 0 0 0 0
0
0 0 0 0 0
0 , 0 0 0 0 0 ,
0
0
1
0
0
3
0
0
2
a35
0,
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
3
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
3
,
,
,
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
,
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9. Does an 8-dimensional vector space contain linear subspaces V1 , V2 , V3 with no common non-zero
element, such that
(a) dimpVi q 5, i 1, 2, 3?
(b) dimpVi q 6, i 1, 2, 3?
(a) We can give an example of this — V1
V3
tp0, 0, p, q, r, s, t, 0qu.
tpa, b, c, d, e, 0, 0, 0qu, V2 tph, k, 0, 0, 0, `, m, nqu and
(b) This can’t happen – since the intersection of a k-dimensional subspace and an `-dimensional
subspace of n is a subspace of dimension at least k ` n, the two six-dimensional subspaces V1
and V2 of 8 intersect in a subspace of dimension at least 4, and so V1 X V2 meets the third subspace
V3 in a subspace of dimension at least 2.
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14. Say you have k linear algebraic equations in n variables; in matrix form we write Ax y. Give
a proof or counterexample for each of the following.
(a) If n k there is always at most one solution.
(b) If n ¡ k you can always solve Ax y.
(c) If n ¡ k the nullspace of A has dimension greater than zero.
(d) If n k then for some y there is no solution of Ax y.
(e) If n k the only solution of Ax 0 is x 0.
(a) This is false. Take
A
Then a vector of the form
1
2
1
2
x
and y 1
t
t
1
2
is a solution for any value of t.
(b) This is also false. Take
A
Then n 3, k
1
2
1
2
1
2
and y 1
1
2 but there is no solution.
(c) This is true, since the dimension of the image of A can be at most k, so the dimension of the
kernel (nullspace) of A is at least n k.
(d) This is true, since the dimension of the image of A is at most n — so for any vector y in the
complement of this (at most) n-dimensional subspace of k there is no solution of Ax y.
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(e) This is false. Take
A
1
2
3
1
2
3
5
Then a vector of the form
x
t
t
solves Ax 0 for any value of t even though n 2 and k
15. Let A :
3.
Rn Ñ Rk be a linear map. Show that the following are equivalent.
(a) A is injective (hence n ¤ k). [injective means one-to-one]
(b) dim kerpAq 0.
(c) A has a left inverse B, so BA I.
(d) The columns of A are linearly independent.
We’ll show (a) ùñ (b) ùñ (c) ùñ (d) ùñ (a).
(a) ùñ (b): Assume A is injective. Then there cannot be a vector other than 0 for which
A0 0, therefore the kernel of A consists only of t0u and so is 0-dimensional.
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(b) ùñ (c): Let te1 , e2 , . . . , en u be a basis of n . Then Ape1 q, Ape2 q, . . . , Apen q are linearly
independent (because if they were not then there would be constants c1 , c2 , . . . , cn , not all zero, for
which
c1 Ape1 q c2 Ape2 q cn pAn q 0.
But then the nonzero vector c1 e1 cn en would be in the kernel of A, contradicting (b)). Now
complete Ape1 q, . . . , Apen q to a basis of k using vectors fn 1 , . . . , fk . Then define B by setting
B pApe1 qq e1 , B pApe2 qq e2 , . . . , B pApen qq en and choosing B pfi q arbitrarily. Then B is a left
inverse for A.
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(c) ùñ (d): The columns of A are the vectors Ape1 q, Ape2 q, . . . , Apen q for the standard basis of
Rn. If they were linearly dependent, then we would have constants c1, c2, . . . , cn so that
c1 Ape1 q
c2 Ape2 q
cn pAn q 0.
But then it would be impossible to have
B pApc1 e1
c2 e2
cn en qq c1 e1
c2 e2
cn en ,
since we would have to have
B pApc1 e1
c2 e2
cn en qq B p0q 0.
Therefore if A has a left inverse, its columns must be linearly independent.
(d) ùñ (a): Since the columns of A are independent, we cannot have
c1 Ape1 q
c2 Ape2 q
cn pAn q 0
for c1 , c2 , . . . , cn not all zero. But if A were not injective there would be two vectors x1 and x2
with x1 x2 and Apx1 q Apx2 q. In other words, the non-zero vector x x1 x2 would satisfy
Apxq 0. Writing x as x c1 e1 cn en shows that this is impossible, so A must be injective.
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16. Let A :
Rn Ñ Rk be a linear map. Show that the following are equivalent.
(a) A is surjective (hence n ¥ k).
(b) dim impAq k.
(c) A has a right inverse B, so AB
(d) The columns of A span
I.
Rk .
Once again, we’ll show (a) ùñ (b) ùñ (c) ùñ (d) ùñ (a).
(a) ùñ (b): Assume A is injective. Then the image of A is all of
Rk , which has dimension k.
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(b) ùñ (c): Since the dimension of impAq is k, every vector in k is in the image of A. Choose a
basis tf1 , f2 , . . . , fk u of k , and then choose vectors e1 , . . . , ek in n such that Ape1 q f1 , Ape2 q f2 ,. . . ,Apek q fk . Set B pf1 q e1 , B pf2 q e2 ,. . . ,B pfk q ek . Then B is a right inverse for A.
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(c) ùñ (d): If A has a right inverse, then for every vector f P k , we have ApB pf qq f . Thus
every vector f P k is in the image of A, which is the span of the columns of A. Thus the columns
of A span k .
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(d) ùñ (a): The span of the columns of A is the image of A. So if the columns of A span all
of k , then every f P k is a linear combination of the columns of A, and the n coefficients in this
linear combination are the components of a vector e P n with the property that Apeq f . Thus
A is surjective.
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19. Consider the system of equations
x
x
y
y
z
2z
a
b.
(a) Find the general solution of the homogeneous equation.
z
(b) A particular solution of the inhomogeneous equations when a 1 and b 2 is x 1, y 1,
1. Find the most general solution of the inhomogeneous equations.
(c) Find some particular solution of the inhomogeneous equations when a 1 and b 2.
(d) Find some particular solution of the inhomogeneous equations when a 3 and b 6.
[Remark: After you have done part (a), it is possible immediately to write the solutions of the
remaining parts.]
(a) We must solve the equations when a b 0, so we row reduce:
1
1
1
1
1
0
2 0
Ñ
1
0
1
2
1
0
3 0
7
From this we see that we can choose z arbitrarily (say z t) and then we will have y
x 12 t. In other words
1 x
2t
y
3
2t
and
.
3
2t
z
t
(b) Since the problem tells us a particular solution, we can immediately conclude
y
1 21 t
1 32 t .
z
1
x
t
(c) Since the right side in this part is the negative of the right side in part (b), by linearity the
particular solution can be chosen to be the negative of the particular solution there, i.e., x 1,
y 1, z 1. So the general solution in this case is
x
y
1 21 t
1 32 t .
1
z
t
(d) Since the right side in this part is 3 times the right side in part (b), by linearity the particular
solution can be chosen to be 3 times the particular solution there, i.e, x 3, y 3, z 3. So the
general solution in this case is
x
3 21 t
y
z
3
3
3
2t
t
22. (a) FInd a 2 2 matrix that rotates the plane by
counterclockwise).
(b) Find a 2 2 matrix that rotates the plane by
horizontal axis.
by
.
45 degrees ( 45 degrees means 45 degrees
45 degrees followed by a reflection across the
(c) Find a 2 2 matrix that reflects across the horizontal axis followed by a rotation of the plane
45 degrees.
(d) Find a matrix that rotates the plane through
60 degrees, keeping the origin fixed.
(e) FInd the inverse of each of these maps.
?
?
?
?
(a) This rotation sends the vector p1, 0q to p1{ 2 , 1{ 2q and sends the vector p0, 1q to p1{ 2 , 1{ 2q.
We put the image vectors in columns and get the matrix
T45
?1
?12
2
?12
?12
.
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(b) Reflection across the horizontal (x) axis sends p1, 0q to itself and sends p0, 1q to p0, 1q and
so its matrix is
1
0
F .
0 1
Therefore rotation followed by reflection is given by the matrix
F T45
1
0
0
1
?1 ?1 ?1 ?1 2
2
2
2
?12
?1 ?1
2
2
?12
(c) On the other hand, reflection followed by rotation is given by
T45 F
(d) Rotation by
?1
1
?
2
2
?12
1
0
?12
60 degrees sends p1, 0q to p1{2 ,
T60
?1
2
0
1
3{2q and sends p0, 1q to
? 23
2
1
2
?
(e) The inverse of T45 is T45 , which sends p1, 0q to p1{ 2 ,
so
T 1 T45 ?12
F ) so we have
pF T45 q1 T451 F 1 T45 F ?1
2
pT45 F q1 F 1 T 1 F T45 45
1
Finally, T60
?
? ?
1{ 2q and sends p0, 1q to p1{ 2 , 1{ 2q,
?12
?1 ?1 2
2
Likewise
1
0
?
p 3{2 , 1{2q. Thus
?1 ?1 2
2
45
Reflection in the x axis is its own inverse (F 1
?
1
2
?3
?12
?12
?12
?12
0
1
1
0
0
1
?1 ?1 2
2
?1 ?1
2
2
?1 ?1 ?1
2
2
2
?12
?12
?12
?12
?12
?
?
T60 which sends p1, 0q to p1{2 , 3{2q and sends p0, 1q to p 3{2 , 1{2q. Thus
1
T60
T60 1
2
?
23
?3 2
1
2
Note that the inverses of the rotation (and reflection) matrices are their transposes. This is not a
coincidence.
23. (a) Find a 3 3 matrix that acts on
x2 x3 plane by 60 degrees.
(b) Find a 3 3 matrix A mapping
the x2 axis fixed.
R3 as follows:
it keeps the x1 axis fixed but rotates the
R3 Ñ R3 that rotes the x1x3 plane by 60 degrees and leaves
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(a) We can use the answer to the preceding?problem for this – apparently
? we want to sent p1, 0, 0q
to itself, and then to send p0, 1, 0q to p0, 1{2 , 3{2q and p0, 0, 1q to p0, 3{2 , 1{2q. So the matrix
is
1 0
0
T
? 23 1
2
0
?3
0
1
2
2
(b) This one is similar, the matrix is
1
2
A
?0
3
2
? 23 0
1
0
1
2
0
24. Find a real 2 2 matrix A (other than A I) such that A5
I.
This continues the “rotation theme” of the last couple of problems. An operation that will
result in the identity map if repeated five times would be rotation by 360{5 72 degrees (or 2π {5
radians). And such a rotation would send p1, 0q to pcosp2π {5q , sinp2π {5qq and would send p0, 1q to
pcospπ{2 2π{5q , sinpπ{2 2π{5q p sinp2π{5q, , cosp2π{5qq. So the matrix is
sinp2π {5q
.
A
cosp2π {5q
Just for the record, the complex number z cosp2π {5q i sinp2π {5q ei2π{5 satisfies the equation
z 5 1 0. Clearly z 1 is a factor of z 5 1 and
z 5 1 pz 1qpz 4 z 3 z 2 z 1q.
cosp2π {5q
sinp2π {5q
To calculate the roots of z 4 z 3 z 2 z 1 we will factor it, and to factor z 4 z 3 z 2 z 1,
we notice that it is a “palindrome”, i.e., its list of coefficients (1,1,1,1,1) reads the same backwards
as forwards. Therefore its roots occur in reciprocal pairs (if α is a root, then so is 1{α — you might
take a moment to prove this). So let’s assume the four roots of z 4 z 3 z 2 z 1 are α, β, 1{α
and 1{β. So we should be able to write
z
4
z
3
z
2
z 1 pz αq z Now let p α
1
α
pzβ q
z
1
β
z
2
α
1
α
z
1
z
2
β
1
β
β 1{β. Then we’ll have
z 4 z 3 z 2 z 1 pz 2 pz 1qpz 2 qz 1q z 4 pp q qz 3 ppq 2qz 2 pp q qz
Therefore p q 1 and pq 2 1, so p and q are roots of the quadratic polynomial
? ?
1
5
1 5
2
2
pz pqpz qq z pp qqz pq z z 1 z z
.
z
1{α and q
2
2
(Ooooh — the golden ratio!) so now we know p and q. This gives us equations for α and β:
α
1
α
1 2
?
5
and β
1
β
1 2
?
5
1.
1
10
?
or
1 5
α
α
2
The quadratic formula gives us
2
10
and β
2
?
1
2
5
1 0.
β
?
?
d ?
d ?
1
51
α 3 5 4 5 1 i 5 5
2
2
2
4
2
2
d ?
? d ?
?
1
1
5
β 3 2 5 4 1 4 5 2i 5 2 5
2
2
and
This gives the other four roots of z 5 1 and since the angle θ 2π {5 is in the first quadrant, we
choose the value of α with the plus sign and get that our matrix is
A
cosp2π {5q
sinp2π {5q
?
?a ? sinp2π{5q 41 p a5 1q 41 2 5 5 .
?
?
?
1
1
cosp2π {5q
5
4 2 5
4 p 5 1q
26. Let L, M , and N be linear maps from the (two dimensional) plane to the plane given in terms
of the standard i, j basis vectors by:
Li j ,
M i i ,
Lj i
(rotation by 90 degrees counterclockwise)
Mj j
(reflection across the vertical axis)
N v v
(reflection across the origin)
(a) Draw pictures describing the actions of the maps L, M , and N and the compositions: LM ,
M L, LN , N L, M N , and N M .
(b) Which pairs of these maps commute?
(c) Which of the following identities are correct — and why?
p1q
p5q
N p2q
2
M I p6q
L2
I p3q
3
M M p7q
N2
L4
I
MNM
N
p4q
p8q
L5
L
NMN
L
(d) Find matrices representing each of the linear maps L, M , and N .
(a) In each case the red F is the image of the blue F under the transformation:
L
M
N
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LM
ML
LN
NL
MN
NM
(b) From the pictures you can see that L does not commute with M , but N commutes with both
L and M .
(c) (1) L2 N is true since rotating by 180 degrees negates a vector, as does reflection through
the origin.
(2) N 2 I is true because double negation brings a vector back to its original position.
(3) L4 I is true because L4 is rotation through 4 90 360 degrees.
(4) L5 L is true by multiplying both sides of equation (3) by L.
(5) M 2 I is true because double reflection brings a vector back to its original position.
(6) M 3 M is true by multiplying both sides of equation (5) by M .
(7) M N M N is true because we know M commutes with N so multiply both sides of M N N M
by M on the right and get M N M N M 2 N I N , using equation (5).
(8) N M N L is false. Apply both sides to the vector i. We have Lpiq j. But N M N piq N M piq N piq i,
(d) The matrices are
L
0
1
31. Think of the matrix A 1
0
a
c
b
d
1
M
0
0
1
,
1 0
N
0 1
.
as mapping one plane to another.
(a) If two lines in the first plane are parallel, show that after being mapped by A they are also
parallel — although they might coincide.
(b) Let Q be the unit square: 0 x 1, 0 y 1 and let Q1 be its image under this map A.
Show that the areapQ1 q |ad bc|. [More generally, the area of any region is magnified by |ad bc|
(ad bc is called the determinant of a 2 2 matrix)].
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(a) The matrix maps the point px, y q to the point pax by , cx dy q. If we write the equation of
a line parametrically as x p1 qt, y r1 st, then the matrix maps the line to the curve given
parametrically by
x app1
y
qtq
cpp1
qtq
bpr1
stq pap1
dpr1
stq pcp1
br1 q
dr1 q
paq
pcq
bsqt
dsqt.
A line parallel to the first one can be parametrized using the same direction vector, so it will be
given by x p2 qt, y r2 st, so it will be mapped to
x pap2
br2 q
paq bsqt
y pcp 2 dr 2q pcq
Since the resulting lines have the same direction vector paq
dsqt
bs , cq
dsq, the mapping sends the
two parallel lines to parallel lines.
(b) The image of the unit square is a parallelogram (by part (a)) with vertices p0, 0q, pa, cq,
b , c dq and pb, dq. To calculate the area of the parallelogram, note that p0, 0q and pa, cq lie
on the line with equation cx ay 0 and pa b , c dq and pb, dq lie on the line with equation
cx ay ? cb ad. The distance between the two lines (the “height” of the parallelogram) is
|cb ad|{ a2 c2 and the “base” of the parallelogram is the distance from p0, 0q to pa, cq, or
?
a2 c2 . So the area is the product of the base and the height, or |bc ad|.
pa
R
R
32. (a) Find a linear map of the plane, A : 2 Ñ 2 that does the following transformation of the
letter F (here the smaller F is transformed to the larger one):
(b) Find a linear map of the plane that inverts this map, that is, it maps the larger F to the
smaller.
13
(a) The vector p1, 0q is mapped to p0, 1q and the vector p0, 5q is mapped to
conclude that p0, 1q is mapped to p2, 0q. Thus the matrix of the map is
p10, 0q so we can
2 0
1
0
(b) The inverse of this map will send p10, 0q to p0, 5q, i.e., it will send p1, 0q to p0, 21 q, and it
will send p0, 1q back to p1, 0q so its matrix will be
0
12
1
0
35. Let A be a matrix, not necessarily square. Say v and w are particular solutions of the equations
Av y1 and Aw y2 , respectively, while z 0 is a solution of the homogeneous equation Az 0.
Answer the following in terms of v, w, and z.
(a) Find some solution of Ax 3y1 .
(b) Find some solution of Ax 5y2 .
(c) Find some solution of Ax 3y1 5y2 .
(d) Find another solution (other than z and 0) of the homogeneous equation Ax 0.
(e) Find two solutions of Ax y1 .
(f) Find another solution of Ax 3y1 5y2 .
(g) If A is a square matrix, then det A ?
(h) If A is a square matrix, for any given vector w can one always find at least one solution of
Ax w? Why?
(a) Since Av y1 , we’ll have Ap3vq 3y1 .
(b) Since Aw y2 , we’ll have Ap5wq 5y2 .
(c) From (a) and (b) we have Ap3v 5wq 3y1 5y2 .
(d) For instance, 2z is another solution of Ap2zq 0.
(e) For instance, Apv
zq y1 and Apv
(f) For instance, Ap3v 5w
2zq y1 .
2zq 3y1 5y2 .
(g) Since there is a non-zero solution of Az 0, we must have det A 0.
(h) No. Since dim kerpAq ¥ 1, we must have dim impAq
vectors w for which there is no solution of Ax w.
¤ n 1 (if A is n n) so there are
14
38. Let L be a 2 2 matrix. For each of the following give a proof or counterexample.
0 then L 0.
(b) If L2 L then either L 0 or L I.
(c) If L2 I then either L I or L I.
(a) If L2
(a) This is false. For instance, choose L (b) This is also false. Choose L 2
will satisfy L
L).
(c) This is false. Choose L 1
0
1
0
0
1
0
0
0
0
1
0
.
(in fact projection onto any one-dimensional subspace
39. Find an example of 2 2 matrices A and B so that AB
For instance, choose A 0
0
1
0
and B
1
0
0
0
0 but BA? 0.
.
43. Let V be the linear space of smooth real-valued functions and L : V
by Lu : u2 u.
ÑV
the linear map defined
(a) Compute Lpe2x q and Lpxq.
(b) Find particular solutions of the inhomogeneous equations
(a) u2
u 7e2x ,
(b) w2
w
4x,
(c z 2
z
7e2x 3x
(c) Find the kernel (nullspace) of L. What is its dimension?
(a) Lpe2x q pe2x q2
5e2x and Lpxq pxq2
(b) (a) Since Lpke2x q 5ke2x we have u 57 e2x .
(b) (b) Since Lpkxq kx, we have w 4x.
(b) (c) Putting these together we’ll have z 75 e2x 3x.
e2x
4e2x
e2x
(c) The kernel of L consists of functions of the form c1 cos x
kernel is 2.
x0
x x.
c2 sin x, so the dimension of the
44. Let PN be the linear space of polynomials of degree at most N and L : PN Ñ PN the linear
map defined by Lu : au2 bu1 cu, where a, b, and c are constants. Assume c 0.
15
(a) Compute Lpxk q.
(b) Show that nullspace (kernel) of L : PN
(c) Show that for every polynomial q pxq
the ODE Lp q.
(d) Find some solution v pxq of v 2
(a) We have Lpxk q apxk q2
v
bpxk q1
Ñ PN is 0.
P PN
there is one and only one solution ppxq
P PN
of
x2 1.
cxk
kpk 1qaxk2
kbxk1
cxk .
(b) Since c 0, L maps any polynomial of degree k (with leading coefficient ak 0 to a
polynomial of degree k (with leading coefficient cak 0). So the only polynomial that L maps to
zero is ppxq 0.
(c) Since the kernel of L is zero on PN , and the dimension of the kernel plus the dimension of
the image is the dimension of PN , we have the dimension of the image is the dimension of PN , so
the image is the entire space PN . Thus L is onto. Since L is one-to-one from part (b), the existence
and uniqueness statement in the problem follows.
(d) We can do this pretty much by trial and error: First, note that Lpx2 q 2 x2 and then
note that Lp1q 1. Put these together to get Lpx2 3q 2 x2 3 x2 1. So a solution (in
fact, the only polynomial solution) is v x2 3.