Download Mathematics 206 Solutions for HWK 13a Section 4.3 p184 Section

Mathematics 206
Solutions for HWK 13a
Section 4.3 p184
Section 4.3 p184 Problem 5. For W the set of all functions that are continuous on [0, 1] and
V the set of all functions that are integrable on [0, 1], verify that W is a subspace of V .
Solution.
We know that continuous functions on [0, 1] are also integrable, so each function
belonging to W does belong to V . In other words, W is at least a subset of V . There are lots of
continuous functions on [0, 1], so W is certainly nonempty. We show that W is both closed under
addition and closed under scalar multiplication.
Closed under addition. Assume that f and g belong to W . In other words, assume that f and
g are continuous on [0, 1]. A theorem of calculus tells us that f + g is continuous at x if f and
g are, so f + g is also continuous on [0, 1]. (This is essentially a consequence of the definition of
continuity together with the fact that the limit of a sum is the sum of the limits, provided those
limits exist.) In other words f + g belongs to W , as required.
Closed under scalar multiplication. Assume that f belongs to W and c is a real number. Let
g = cf . Here, too, a theorem of calculus tells us that if f is continuous at x, then so is the function
we’ve called g. Therefore g belongs to W , as required.
Section 4.3 p184 Problem 9. For W the set of all nonnegative functions in C(−∞, ∞), give
a specific example to show that W fails the subspace test and is therefore not a vector subspace of
C(−∞, ∞).
Solution. Although W is both nonempty and closed under addition, it fails to be closed under
scalar multiplication. One easy example that could be used to show this is to let f be, for instance,
either the constant function 1 or the function that takes output to be the square of input (i.e.
f (x) = x2 ) and take c to be a negative real number, for instance c = −2. Then we have f ∈ W
but cf ∈
/ W.
Section 4.3 p184 Problem 11. For W the set of all n × n matrices with zero determinant, give
a specific example to show that W fails the subspace test and is therefore not a vector subspace of
Mn,n.
Solution.
In this case, W is closed under scalar multiplication but fails to be closed under
addition. Once n is specified,
easy to
· it’s pretty
¸
· cook¸up an example to show this. With n = 2, for
0 0
2 1
instance we could put A =
and B =
. Then |A| = |B| = 0 but |A + B| = 1 6= 0.
1 1
0 0
In other words, A ∈ W, B ∈ W, but A + B ∈
/ W.
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A. Sontag
March 25, 2001
Math 206 HWK 13a Solns contd
4.3 p184
Section 4.3 p184 Problem 13. Which of the following subsets of C(−∞, ∞) are subspaces of
C(−∞, ∞)?
(a) The set of all nonnegative functions (i.e. the set of all functions f satisfying f (x) ≥ 0 for every
x). This set is not closed under scalar multiplication, hence NOT a subspace.
(b) The set of all even functions (i.e. the set of all functions f satisfying f (−x) = −f (x) for every
x) is a subspace. [Proof. We know even functions exist. Suppose f and g are even and c is a real
number. Then, for every x, we have (fg )(−x) = f (−x) + g(−x) = f (x) + g(x) = (f + g)(x) and
(cf )(−x) = c(f (−x)) = c(f (x)) = (cf )(x). ]
(c) The set of all odd functions (i.e. the set of all functions f satisfying f (−x) = −f (x) for every
x) is a subspace. The reasoning is similar to that for (b).
(d) The set of all constant functions is a subspace. Constant functions exist, the sum of two
constant functions is also constant, and every scalar multiple of a constant function is a constant
function.
(e) The set, W , say, of all functions f such that f (0) = 0. This set is a subspace. The constant
function zero satisfies the required condition and therefore belongs to W . If f, g ∈ W , then
(f + g)(0) = f (0) + g(0) = 0 + 0 = 0 so f + G ∈ W . Similarly, if f ∈ W and c is a scalar, then
(cf )(0) = c(f (0) = c(0) = 0 so the function cf also belongs to W .
(f) The set, W , say, of all functions f such that f (0) = 1. This set is NOT a subspace. Although
it’s nonempty (the constant function 1 belongs), it is neither closed under addition nor closed under
scalar multiplication. For instance, if f, g ∈ W , then (f + g)(0) = f (0) + g(0) = 1 + 1 = 2 6= 1 so
f +g ∈
/ W.
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A. Sontag
March 25, 2001
Math 206 HWK 13a Solns contd
4.3 p184
Section 4.3 p184 Problem 17. Given W = {(a, b, a + 2b) : a, b ∈ R}, decide whether W is a
vector subspace of R3 (assuming the usual operations).
Solution. Another way to describe W is as follows: W = {(x, y, z) ∈ R3 : z = x + 2y}. It’s easy
to see that (0, 0, 0) ∈ W . If (x, y, z) ∈ W and (x0 , y 0 , z 0 ) ∈ W , then we have (x, y, z) + (x0 , y 0 , z 0 ) =
(x+x0 , y+y 0 , z+z 0 ) and z+z 0 = (x+2y)+(x0 +2y 0 ) = (x+x0 )+2(y+y 0 ), so (x, y, z)+(x0 , y 0 , z 0 ) ∈ W .
In other words, W is closed under addition. Similarly, if (x, y, z) ∈ W and c is a scalar, then
c(x, y, z) = (cx, cy, cz) and cz = c(x + 2y) = cx + 2(cy) so c(x, y, z) ∈ W . In other words, W is
also closed under scalar multiplication. Conclusion: yes, W is a subspace of R3 .
Section 4.3 p184 Problem 19. Given W = {(x1 , x2 , x1 x2 ) : x1 , x2 ∈ R}, decide whether W is
a vector subspace of R3 (assuming the usual operations).
Solution. We have W = {(x, y, z) ∈ R3 : z = xy}. Although W is obviously nonempty, it is
neither closed under addition nor closed under scalar multiplication. For instance, (1, 1, 1) ∈ W
but (1, 1, 1) + (1, 1, 1) = (2, 2, 2) ∈
/ W and 2(1, 1, 1) = (2, 2, 2) ∈
/ W.
Section 4.3 p184 Problem 23. (Guided Proof.) Let W be a nonempty subset W of a vector
space V . Prove that W is a subspace of V iff ax + by ∈ W for all scalars a and b and all vectors
x, y ∈ W .
Proof. (=⇒). Assume that W is a subspace of V .
Then assume that x, y ∈ W and a, b ∈ R.
As a subspace, W is closed under scalar multiplication, so ax ∈ W and by ∈ W .
Then, since W is also closed under addition, we see that ax + by ∈ W , as required.
(⇐=). Now assume, for all x, y ∈ W and all scalars a, b, that ax + by ∈ W .
To show that W is a subspace, it suffices to show (we already know W is nonempty) that W is
both closed under addition and closed under scalar multiplication.
Assume x, y ∈ W . Applying our hypothesis with this same x and y and with a = 1, b = 1, we see
that x + y must belong to W . Since x and y were chosen arbitrarily, it follows that W is closed
under addition.
Finally, assume that x ∈ W and c is a scalar. Applying our hypothesis with x as chosen, y
unspecified, a = c and b = 0, we see that ax + by = cx + 0 = cx must belong to W . Therefore W
is also closed under scalar multiplication.
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A. Sontag
March 25, 2001
Math 206 HWK 13a Solns contd
4.3 p184
· ¸
1
Section 4.3 p184 Problem 25. Let A be a fixed 2×3 matrix. Let W := {x ∈ R : Ax =
}.
2
Prove that W is not a subspace of R3 .
3
· ¸
· ¸
2
1
Proof. Assume that x ∈ W . Then A(2x) = 2Ax =
6=
} so 2x ∈
/ W . Since W is not
4
2
closed under scalar multiplication, it can’t be a subspace. (Alternatively, you could use a similar
argument to show that W is not closed under addition. Note that W might or might not be
nonempty, depending on the particular matrix A. A more general result is that the solution set
for a nonhomogeneous system of linear equations will NEVER be a vector space under the usual
operations.)
Section 4.3 p184 Problem 26. Let A be a fixed m × n matrix. Let W := {x ∈ Rn : Ax = 0}.
Prove that W is a subspace of Rn .
Proof. The zero vector for Rn obviously belongs to W . Suppose that x, y ∈ W and let c be a
scalar. Then
A(x + y) = Ax + Ay = 0 + 0 = 0
and
A(cx) = cAx = c0 = 0
so x + y and cx both belong to W . We’ve shown that W is nonempty, closed under addition, and
closed under scalar multiplication. Therefore W is a subspace. (So what you’ve just shown is that
the solution set for a homogenous system of linear equations is ALWAYS a vector space, under the
usual operations.)
Section 4.3 p184 Problem 29. Let A be a fixed 2 × 2 matrix. Let W := {X ∈ M2,2 : XA =
AX} is a subspace of M2,2 . (W is sometimes referred to as the set of 2 × 2 matrices that commute
with the given matrix A.) Prove that W is a subspace of M2,2 .
Proof. It is easy to see that the matrix 02,2 belongs to W , since 02,2 A = 02,2 = A02,2 .
Suppose X, Y ∈ W . The distributivity of matrix multiplication give us (X + Y )A = XA + Y A and
A(X + Y ) = AX + AY . The hypothesis on X and Y gives AX = XA and AY = Y A. Therefore
(X + Y )A = XA + Y A = AX + AY = A(X + Y ), which shows that X + Y belongs to W , as
required.
Now suppose that X ∈ W as before and that c is a scalar. Then A(cX) = c(AX) = c(XA) =
(cX)A. This shows that cX must belong to W .
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A. Sontag
March 25, 2001