Download Practice session 3 (week 3) — Solutions

T HE U NIVERSITY OF S YDNEY
P URE M ATHEMATICS
Linear Mathematics
2009
Practice session 3 (week 3) — Solutions
1.
n³ ´ ³ ´ ³ ´o
3
1
4
0 , 2 , 1
a) Let X =
. Show that X is a linearly independent subset of R3 .
0
0
2
b) Let Y = {f1 (x), f2 (x), f3 (x)}, where f1 (x) = sin x, f2 (x) = sin 3x and f3 (x) = sin 5x.
Show that Y is a linearly independent subset of F.
Solution
³ ´
³ ´
³ ´ ³ ´
3
1
4
0
a) Suppose a 0 + b 2 + c 1 = 0 . Then
0
0
2
0
3a + xb + 4c = 0
2b + c = 0
2c = 0
By inspection there is only one solution; namely, a = b = c = 0. Hence, X is linearly
independent.
b) Suppose a sin x + b sin 3x + c sin 5x = 0, for all x. Putting x = π2 , x = π3 and x = π4 ,
respectively, gives the following three equations:
√
a −
3
a
2
√1 a
2
b +
−
1
√
+
b −
2
√
c = 0,
= 0,
= 0,
(x = 0),
(x = π2 ),
(x = π3 ).
3
c
2
√1 c
2
√
Taking out the common factor of 23 from the second equation and the common factor of
√1 from the second equation the augmented matrix for this system of equations is
2
à 1
1
1
2.
−1
1
0 −1
1 −1
0 !
0 .
0
Using Gaussian elimination the only solution for this system of equations is a = b = c =
0, so Y is linearly independent as claimed.
³³ ´ ³ ´ ³ ´´
4
1
3
0 , 2 , 1
. Determine dim V and whether or not V = R3 .
a) Let V = Span
2
0
0
¡
¢
b) Let W = Span f1 (x), f2 (x), f3 (x) . Determine dim W and whether or not W = F.
Solution
a) By Question 1(a), these three vectors are linearly independent so dim V = 3. As V is a
three dimensional subspace of and dim R3 = 3 we must have V = R3 .
b) By definition, W is the subspace of F spanned by {f1 , f2 , f3 }. In Question 1(d) we showed
that this set of vectors in linearly independent, so {f1 , f2 , f3 } forms a basis for W and
dim W = 3. However, W 6= F since if n ≥ 0 then Pn ⊂ F, forcing F to be infinite
dimensional. Hence, W 6= F.
Math 2061: Practice session 3 (week 3) — Solutions
S.B and A.M 2/1/2009
Linear Mathematics
Practice session 3 (week 3) — Solutions
Page 2
3. Let f1 (x) = cos x, f2 (x) = cos(x + 1) and f3 (x) = sin x. Show that {f1 , f2 , f3 } is a linearly
dependent subset of F and find the dimension of Span(f1 , f2 , f3 ).
Solution Suppose af1 + bf2 + cf3 = 0 in F. (Note that 0 is the zero function in F; that is, 0 is
the function 0(x) = 0, for all x ∈ R. Then a cos x + b cos(x + 1) + c sin x = 0, for all x ∈ R.
Now, using the double angle formula, cos(x + 1) = cos x cos 1 − sin x sin 1, that is,
(cos 1) cos x − 1 cos(x + 1) − (sin 1) sin x = 0,
for all x ∈ R. Thus, we have a non–zero linear combination of f1 , f2 and f3 which sums to 0.
Hence, {f1 , f2 , f3 } is linearly dependent.
To find the dimension of V = Span(f1 , f2 , f3 ) we must first find a basis of this subspace.
By the last paragraph, Span(f1 , f2 , f3 ) = Span(f1 , f3 ) since f2 is a linear combination of f1
and f3 . On the other hand, f1 is not a scalar multiple of f3 , so {f1 , f3 } is linearly independent.
Therefore, {f1 , f3 } is a basis of V and dim V = 2.
¯
o
n³w´
x
4¯
4. Find a basis of the subspace V =
∈ R ¯ 2x − y = z − 3w of R4 .
y
z
³w´
Solution A vector xy ∈ R4 belongs to V if and only if 2x − y = z − 3w or, equivalently, if
z
³w´
y = 2x − z + 3w. Thus, an arbitrary element xy of V can be written in the form
z
³w´
x
y
z
³
=
´
w
x
2x−z+3w
z
=w
³1´
0
3
0
+x
³0´
1
2
0
³
+z
0 ´
0
−1 .
1
³1´ ³0´
³ 0 ´
0
0
1
Therefore, the vectors 3 , 2 and −1
span V . We now test this set for linear indepen0
0
³1´
³0´
³ 01 ´ ³ 0 ´
0
dence. Suppose that a 03 + b 12 + c −1
= 00 , for some a, b, c ∈ R. The corresponding
0
0
0
1
augmented matrix, and a row echelon form for this matrix, is

1 0
 0 1

 3 2
0 0
0
0
−1
1

0
Row reduce
0 
−
−−−−→
0 
0

1
 0

 0
0
0
1
0
0
0
0
1
0

0
0 
.
0 
0
Hence, a = b = c = 0 so these three vectors are linearly independent. (Note that there are
no free variables. The last line of the row echelon form says that this system of equations is
consistent.) Therefore, V is a three dimensional subspace of R4 with basis
n³ 1 ´ ³ 0 ´ ³ 0 ´o
0
0 , 1 ,
.
−1
2
3
0
0
1
5. Let X = {p1 , p2 , p3 }, where p1 (x) = x2 − 1, p2 (x) = x(x − 1), p3 (x) = x(x + 1).
a) Show that X is a basis of P2 .
b) Find the unique expression for p(x) = 9x2 − x − 4 as a linear combination of vectors in X.
Solution
a) The subset X of P2 is a basis of P2 if X is linearly independent and Span(X) = P2 .
However, as we will explain, to show that X is a basis of P2 it is enough to either show
that X is linearly independent, or show that Span(X) = P2
Math 2061: Practice session 3 (week 3) — Solutions
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Linear Mathematics
Practice session 3 (week 3) — Solutions
Page 3
Solution 1
We show that X is linearly independent. Suppose that ap1 (x) + bp2 (x) + cp3 (x) = 0, for
some a, b, c ∈ R. That is,
a(x2 − 1) + b(x2 − x) + c(x2 + x) = 0.
Looking at the coefficients of the constant term, the degree 1 terms and the degree two
terms on both sides of this equation we obtain the following system of equations:
−a
= 0,
− b + c = 0,
a + b + c = 0.
The corresponding augmented matrix and row echelon form is therefore,
à −1
0
1
à 1 0
0 !
0
Row reduce
0
−−−−−→ 0 1 −1
0 0
1
0
0 0
−1 1
1 1
0 !
0 .
0
Hence, we have that a = b = c = 0. Therefore, the polynomial functions p1 (x), p2 (x) and
p3 (x) are linearly independent elements of P2 . Therefore, Span(X) is a three dimensional
subspace of P2 . However, dim P2 = 3, so we must have that Span(X) = P2 (since if
W 6= V is a subspace of V then dim W < dim V ). Hence, X is a basis of P2 .
Solution 2
We show that Span(X) = P − 2. Let α + βx + γx2 be an arbitrary element of P2 . Then
α + βx + γx2 ∈ Span(X) if and only if we can find a, b, c ∈ R such that α + βx + γx2 =
ap1 (x)+bp2 (x)+cp3 (x). As above, this is equivalent to the following system of equations
−a
= α,
− b + c = β,
a + b + c = γ,
and the corresponding augmented matrix and row echelon form for this system are, respectively,
à −1
0 0
0 −1 1
1
1 1
à 1 0
0
α !
Row reduce
β
−−−−−→ 0 1 −1
γ
0 0
1
−α !
−β .
1
(α
+
β
+
γ)
2
(Check!) Hence, we can solve for a, b, c. Therefore, Span(X) = P2 . This also shows
that X must be linearly independent because otherwise some subset of X would span P2
which would force dim P2 < 3, a contraction. (Alternatively, set α = β = γ = 0 in the
calculation above and note a, b and c must all be zero.)
b) In Solution 2 above we showed how to write an arbitrary polynomial in P2 as a linear
combination of the polynomials p1 (x), p2 (x), p3 (x). Using back substitution in the row
echelon matrix at the end of Solution 2 we find that, for any α, β, γ ∈ R,
¡
¢
α + βx + γx2 = 12 (α + β + γ)p3 (x) + 12 (α + β + γ) − β p2 (x) − αp1 (x)
= 12 (α + β + γ)p3 (x) + 12 (α − β + γ)p2 (x) − αp1 (x).
Hence, in particular, 9x2 − x − 4 = 4p1 (x) + 3p2 (x) + 2p3 (x).
Math 2061: Practice session 3 (week 3) — Solutions
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