Description

Dane Johnson
Exercise 1
Consider the space V = {v = (v1, …,vn) ∈ Rn | v = ∇f(0) for some function f ∈
C1(Rn) defined in a neighborhood of the origin}
(1) Prove that V , equipped with the usual vector sum and scalar multiplication operations, is a vector space.
To prove this statement, we verify that all the vector space axioms hold. (Note: I realize that much effort could be saved if we first just notice that V ⊆ Rn and then verify that V is closed under vector addition and multiplication by a scalar but perhaps this was not the intent of the exercise and I also didn’t notice this until after checking all vector space axioms manually). Proof
Let u,v,w ∈ V . By definition of the set V there exist functions f,g,h : Rn → R with each function continuously differentiable and defined in a neighborhood of the origin such that u = ∇f(0),v = ∇g(0), and w = ∇h(0).
VS 1
(u + v) + w = (∇f(0) + ∇g(0)) + ∇h(0)
= [(∂1f(0),…,∂nf(0)) + (∂1g(0),…,∂ng(0))] + (∂1h(0),…,∂nh(0))
= [(∂1f(0) + ∂1g(0),…,∂nf(0) + ∂ng(0))] + (∂1h(0),…,∂nh(0))
= (∂1f(0) + ∂1g(0),…,∂nf(0) + ∂ng(0)) + (∂1h(0),…,∂nh(0))
= (∂1f(0) + ∂1g(0) + ∂1h(0),…,∂nf(0) + ∂ng(0) + ∂nh(0))
= (∂1f(0) + (∂1g(0) + ∂1h(0)),…,∂nf(0) + (∂ng(0) + ∂nh(0)))
= (∂1f(0),…,∂nf(0)) + (∂1g(0) + ∂1h(0),…,∂ng(0) + ∂nh(0))
= (∂1f(0),…,∂nf(0)) + [(∂1g(0),…,∂ng(0)) + (∂1h(0),…,∂nh(0))]
= ∇f(0) + (∇g(0) + ∇h(0)) = u + (v + w).
VS 2
Let i : Rn → R, i(x1,…,xn) = 0. Then ∇i = (0,….,0) and in particular ∇i(0) = (0,…,0). Define the vector O = (0,…,0) ∈ Rn. Then it is the case that O = ∇i(0), so that O ∈ V . We claim that O is the additive identity of
V . To see this note that
u + O = (∂1f(0),…,∂nf(0)) + (0,…,0)
= (∂1f(0) + 0,…,∂nf(0) + 0)
= (0 + ∂1f(0),…,0 + ∂nf(0)) = O + u
O + u = (0 + ∂1f(0),…,0 + ∂nf(0)) = (∂1f(0),…,∂nf(0)) = u.
Therefore O ∈ V and since u was arbitrary O + u = u + O = u for any choice of u ∈ V .
VS 3
Given u = ∇f(0), let −u = −∇f(0). Then it is the case that −u ∈ V and we have
u+(−u) = ∇f(0)−∇f(0) = (∂1f(0)−∂1f(0),…,∂nf(0)−∂nf(0)) = (0,…,0) = O.
VS 4
Given u,v ∈ V , with u = ∇f(0),v = ∇g(0), we have
u + v = (∂1f(0),…,∂nf(0)) + (∂1g(0),…,∂ng(0))
= (∂1f(0) + ∂1g(0),…,∂nf(0) + ∂ng(0))
= (∂1g(0) + ∂1f(0),…,∂ng(0) + ∂nf(0))
= v + u.
VS 5
Let c be a scalar (we assume the field in this exercise is the field of real numbers). Then
c[u + v] = c[(∂1f(0) + ∂1g(0),…,∂nf(0) + ∂ng(0))]
= (c∂1f(0) + c∂1g(0),…,c∂nf(0) + c∂ng(0))
= (c∂1f(0),…,c∂nf(0)) + (c∂1g(0),…,c∂ng(0))
= c(∂1f(0),…,∂nf(0)) + c(∂1g(0),…,∂ng(0))
= cu + cv .
VS 6
If a and b are two numbers, then
(a + b)u = ((a + b)∂1f(0),…,(a + b)∂nf(0))
= (a∂1f(0) + b∂1f(0),…,a∂nf(0) + b∂nf(0))
= (a∂1f(0),…,a∂nf(0)) + (b∂1f(0),…,b∂nf(0))
= a(∂1f(0),…,∂nf(0)) + b(∂1f(0),…,∂nf(0))
= au + bu.
VS 7
If a and b are two numbers, then
(ab)u = ((ab)∂1f(0),…,(ab)∂nf(0))
= (a(b∂1f(0)),…,a(b∂nf(0)))
= a(b∂1f(0),…,b∂nf(0)) = a(bu).
VS 8
Using the arbitrary element u ∈ V , we have
1u = 1(∂1f(0),…,∂nf(0)) = (1∂1f(0),…,1∂nf(0)) = (∂1f(0),…,∂nf(0)) = u We conclude that the space V is a vector space.
(2) Prove that V = Rn.
Let v ∈ V . There exists a function f : Rn → R such that v = ∇f(0) =
(0)). Since f maps from Rn to R, it is
also the case that . But this means that when evaluating the partial derivatives we have so that
. Therefore, V ⊆ Rn.
Let P = (c1,c2,…,cn) ∈ Rn. Consider the function g : Rn → R given by the rule g(x1,x2,…,xn) = c1x1 + c2x2 + … + cnxn. Then we have ∇g = (c1,…,cn), which means ∇g(0) = (c1,…,cn) as well. If we define w = ∇g(0) we see that since g is defined in any neighborhood of zero and is continuously differentialable that P = (c1,…,cn) = w ∈ V . Therefore
Rn ⊆ V .
Since V ⊆ Rn and Rn ⊆ V we conclude that V = Rn.
Exercise 2
Let V be a vector space over the field K. If X and Y are subspaces of V, then the intersection X ∩ Y is also a subspace of V.
Proof: To prove that X ∩ Y is a subspace of V, it suffices to show that X ∩ Y is a subset of V that is closed under scalar multiplication and vector addition.
Since X ∩ Y ⊆ X and X ⊆ V, we have X ∩ Y ⊆ V.
Let u,v ∈ X∩Y. Then u,v ∈ X. Since X is a subspace of V, X is itself a vector space, so that it must be the case that u+v ∈ X. Similarly, we know that u,v ∈ Y and that Y is also a vector space. Thus, u + v ∈ Y. We have therefore shown that u + v ∈ X ∩ Y.
Since X ∩ Y is a subset of V and is closed under vector addition and scalar multiplication, we conclude that X ∩ Y is a subspace of V.
Note: The textbook mentions that we must show that X ∩ Y contains the zero vector of the space V but this actually follows from the fact that X ∩ Y is closed under scalar multiplication (use the identity element of the field K as the scalar).
Exercise 3
Consider
X = {x = (x1,x2,x3) | a1x1 + a2x2 + a3x3 = 0}
Y = {x = (x1,x2,x3) | b1x1 + b2x2 + b3x3 = 0}
where ai,bi ∈ R for i = 1,2,3 .
(1)
To prove that X and Y are vector spaces we just prove the case for X since the proof for Y is identical if we replace the 0a0 coefficients with 0b0 coefficients.
Instead of verifying each vector space axiom, we instead first note that X ⊆ R3. Since R3 is already known to be a vector space, it suffices to show that X is closed under vector addition and scalar multiplication (with X inheriting the usual operations for the vector space R3).
Let x,y ∈ X with x = (a1x1,a2x2,a3x3) and y = (a1y1,a2y2,a3y3) and let k ∈ R. We have
x + y = (a1x1 + a1y1,a2x2 + a2y2,a3x3 + a3y3) .
To see that x + y ∈ X, note that
a1x1+a1y1+a2x2+a2y2+a3x3+a3y3 = a1x1+a2x2+a3x3+a1y1+a2y2+a3y3 = 0+0 = 0 .
Consider next kx = (kc1x1,kc2x2,kc3x3). Since kc1x1+kc2x2+kc3x3 = k(c1x1 + c2x2 + c3x3) = k(0) = 0, we know kx ∈ X.
Since X is closed under scalar multiplication and vector addition, X is a subspace of R3 and therefore a vector space.
(2)
There are several geometric possibilities for X ∩ Y considering that although we know a1,a2,a3,b1,b2, and b3 are real numbers we have not specified exactly which real numbers they are.
In the case that both X and Y are planes, then by their definition they must both pass through the origin. If ai = bi for each i, then these two planes are actually the same plane so that the intersection is also a plane. If the two planes are distinct then the intersection is a line in R3.
We conclude that there are three possible geometric representations of X∩Y: a line passing through the origin, a plane passing through the origin, or all of R3.
Exercise 4
(1)
The set X = {x ∈ Rn | Ax = 0}, where A is a given m × n matrix, is a subspace of the vector space Rn.
Let x,y ∈ X, so that Ax = Ay = 0. Then we have A(x+y) = Ax+Ay = 0 + 0 = 0, which shows that x + y ∈ X.
Let c ∈ R. We have A(cx) = c(Ax) = c(0) = 0, which shows that cx ∈ X. Thus X is a subspace of Rn.
(2)
The set X = {p ∈ P | p(x) = p(−x) for all x ∈ R} is is a subspace of the vector space of all polynomials with real coefficients P.
Let p,q ∈ X. Then (p + q)(x) = p(x) + q(x) = p(−x) + q(−x) = (p + q)(−x). Thus p + q ∈ X.
Let k ∈ R. Then kp(x) = kp(−x), which shows that if p ∈ X it is also the case that kp ∈ X. Thus X is a subspace of P.
(3)
The set X = {p ∈ P | p has degree less than or equal to n} is a subspace of P.
Let p,q ∈ X. Then p and q are polynomials of degree at most n. The result of summing of two polynomials of degree at most n (to do this perform the real number sums of the corresponding coefficients of each polynomial as usual) must also be a polynomial of degree at most n. Therefore, p+q ∈ X.
Let k ∈ R. Then since the degree of p ∈ X is at most n, the polynomial kp is of degree at most n (note that if k = 0, then kp is the zero polynomial, but this is still of degree less than n no matter which of the common conventions we use for defining the degree of the zero polynomial – but it is impossible for kp to be of higher degree than p). Thus kp ∈ X. Therefore X is subspace of P.
(4)
The set X = {f ∈ C[0,1] | f(1) = 2f(0)} is a subspace of C[0,1], where C[0,1] is the set of all continuous functions on [0,1].
Let f,g ∈ X, which means that f and g are continuous and that f(1) =
2f(0),g(1) = 2g(0). Then we have
(f + g)(1) = f(1) + g(1) = 2f(0) + 2g(0) = 2(f + g)(0) .
This shows that f + g ∈ X
Let k ∈ R. Then for f ∈ X, since f(1) = 2f(0) it follows that kf(1) = k[2f(0)] = 2kf(0). Thus kf ∈ X as well. Therefore X is a subspace of C[0,1].
(5)
The unit sphere in Rn is not a subspace of Rn. Since the claim that the unit sphere in Rn is a subspace of Rn in the general case (that is, for any choice of n), it suffices to give a counterexample for n = 2. This counterexample is amenable to generalization if desired.
Let (x1,y1),(x2,y2) ∈ U, where U is the unit sphere in R2. This means
= 1 and = 1. We show that U is not closed under vector
addition. We define vector addition in the usual way, giving
(x1,y1) + (x2,y2) = (x1 + x2,y1 + y2) .
However, we see that
.
For (x1,y1)+(x2,y2) ∈ U, we would need 2+2x1x2 +2y1y2 = 1. To see why this does not hold generally, take (x1,y1) = (1,0) and (x2,y2) = (0,1). Then (1,0) + (0,1) = (1,1) ∈/ U.
Exercise 5
(1)
(2)
Let M be the set of all n × n symmetric matrices with real entries with proposed field also the set R. Then M is a subspace of the vector space of all n×n real matrices. The dimension of M is 6 and a basis for the subspace
is
.
Since it is not required by the exercise to rigorously prove these statements and the proof that this is a basis will be very tedious, the proof is omitted. (3)