Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 91
Subsection Three.II.2: Rangespace and Nullspace
Three.II.2.22 First, to answer whether a polynomial is in the nullspace, we have to consider it as a
member of the domain P 3 . To answer whether it is in the rangespace, we consider it as a member of
the codomain P 4 . That is, for p(x) = x 4 , the question of whether it is in the rangespace is sensible but
the question of whether it is in the nullspace is not because it is not even in the domain.
(a) The polynomial x 3 ∈ P 3 is not in the nullspace because h(x 3 ) = x 4 is not the zero polynomial in
P 4 . The polynomial x 3 ∈ P 4 is in the rangespace because x 2 ∈ P 3 is mapped by h to x 3 .
(b) The answer to both questions is, “Yes, because h(0) = 0.” The polynomial 0 ∈ P 3 is in the
nullspace because it is mapped by h to the zero polynomial in P 4 . The polynomial 0 ∈ P 4 is in the
rangespace because it is the image, under h, of 0 ∈ P 3 .
(c) The polynomial 7 ∈ P 3 is not in the nullspace because h(7) = 7x is not the zero polynomial in
P 4 . The polynomial x 3 ∈ P 4 is not in the rangespace because there is no member of the domain
that when multiplied by x gives the constant polynomial p(x) = 7.
(d) The polynomial 12x − 0.5x 3 ∈ P 3 is not in the nullspace because h(12x − 0.5x 3 ) = 12x 2 − 0.5x 4 .
The polynomial 12x − 0.5x 3 ∈ P 4 is in the rangespace because it is the image of 12 − 0.5x 2 .
(e) The polynomial 1 + 3x 2 − x 3 ∈ P 3 is not in the nullspace because h(1 + 3x 2 − x 3 ) = x + 3x 3 − x 4 .
The polynomial 1 + 3x 2 − x 3 ∈ P 4 is not in the rangespace because of the constant term.
Three.II.2.23 (a) The nullspace is
(
a
N (h) = { ∈ R
b)
∣ ( ) 2 0 ∣∣
a + ax + ax 2 + 0x 3 = 0 + 0x + 0x 2 + 0x 3 } = { b ∈ R}
b
while the rangespace is
∣
R(h) = {a + ax + ax 2 ∈ P 3 a, b ∈ R} = {a · (1 + x + x 2 ) ∣ a ∈ R}
and so the nullity is one and the rank is one.
(b) The nullspace is this.
( ) ( ) a b ∣∣ −d b ∣∣
N (h) = { a + d = 0} = {
b, c, d ∈ R}
c d
c d
The rangespace
R(h){a + d ∣ a, b, c, d ∈ R}
is all of R (we can get any real number by taking d to be 0 and taking a to be the desired number).
Thus, the nullity is three and the rank is one.
(c) The nullspace is
( ) ( )
a b ∣∣ −b − c b ∣∣
N (h) = { a + b + c = 0 and d = 0} = {
b, c ∈ R}
c d
c 0
while the rangespace is R(h) = {r + sx 2 ∣ ∣ r, s ∈ R}. Thus, the nullity is two and the rank is two.
(d) The nullspace is all of R 3 so the nullity is three. The rangespace is the trivial subspace of R 4 so
the rank is zero.
Three.II.2.24 For each, use the result that the rank plus the nullity equals the dimension of the
domain.
(a) 0 (b) 3 (c) 3 (d) 0
Three.II.2.25 Because
d
dx (a 0 + a 1 x + · · · + a n x n ) = a 1 + 2a 2 x + 3a 3 x 2 + · · · + na n x n−1
we have this.
N ( d
dx ) = {a 0 + · · · + a n x ∣ n a 1 + 2a 2 x + · · · + na n x n−1 = 0 + 0x + · · · + 0x n−1 }
= {a 0 + · · · + a n x ∣ n a 1 = 0, and a 2 = 0, . . . , a n = 0}
In the same way,
for k ≤ n.
= {a 0 + 0x + 0x 2 + · · · + 0x n ∣ ∣ a 0 ∈ R}
N ( dk
dx k ) = {a 0 + a 1 x + · · · + a n x n ∣ ∣ a 0 , . . . , a k−1 ∈ R}