or alternatively
Given this form, we see that in order for a solution to exist for any e,f ∈ F then a,b,c,d need to be such that the vectors
spans the entire space Fcols2, where Fcols2 denotes the vector space of two dimensional column-vectors.
such that vi = (a1i,…,ani) with each aij ∈ F and Aij = aij.
Given this definition of A, we see that
that is, v1t,…,vnt are the columns of A.
+ cmvm = b. Since v1,…,vm,b are all
row vectors, this is akin to saying that
but this is just Act = bt where ct = (c1,…,cm)t. Hence for any vector bt in the column-space of Fn, Ax = bt will have a solution, i.e. the row-reduced echelon form of A will have a pivot in each row, so it will have n pivots.
+ xmvmt = b. This is equivalent to saying that for any b′∈ Fn there exist c1,…,cm ∈ F such
that
Hence v1,…,vm span Fn.
.
defined by {A⋅B : B ∈ Mn×k
}. For b,b′∈ F and C,C′∈ S there exist B,B′∈ Mm×k
such that both C = AB and C′ = AB′. Making use of this and Lemma A.1 we attain
and thus bC + b′C′∈ S. Given this and that 0M
m×k
are both in S, so S is a subspace of Mn×k
.
defined by {B ⋅A : B ∈ Mk×m
}. For b,b′∈ F and C,C′∈ S there exists B,B′∈ Mk×m
such that C = BA and C′ = B′A. Making use of this we obtain the following.
Therefore bC + b′C′∈ S. Since this is true and 0M
k×n
∈ S, then S is a subspace of Mk×n
.
defined by {B ∈ Mn×k(F) : A ⋅ B = 0}. For b,b′∈ F and B,B′∈ S we
have
since AB = 0 and AB′ = 0. This implies bB + b′B′∈ S, so S is a subspace of Mn×k
.
defined by {B ∈ Mk×m(F) : B ⋅ A = 0}. For b,b′∈ F and B,B′∈ S we
have
by Lemma A.1 and because BA = 0 and B′A = 0. This indicates that bB + b′B′∈ S. Thus S is a subspace of Mk×m(F)
which is an element of W1.
So for a,b ∈ F we have the following, given that a(x,y,z) + b(x′,y′,z′) = (ax + bx′,ay + by′,az + bz′)
then a(x,y,z) + b(x′,y′,z′) ∈ W3, and thus W3 a subspace of ℝ3.
So for a,b ∈ F we have the following, given that a(x,y,z) + b(x′,y′,z′) = (ax + bx′,ay + by′,az + bz′)
then a(x,y,z) + b(x′,y′,z′) ∈ W4, and hence W4 is a subspace of ℝ3
So (3,
,1) and (3,
,2) are both in W6. They’re sum, however, is (6,
+ 
,3) which since
then W6 is not closed under addition, and so its not a subspace of ℝ3.
for which the entries of A are all zero except for any one entry. There will only
be mn of these matrices in Mm×n
, and they will span the space.
More formally, we can define S as the set {Aℓk : Aℓk ∈ Mm×n
,k ∈{1,…,m},ℓ ∈{1,…,n}} where each entry of the
matrix Aℓk is defined by
where δ is the Kronecker-delta according to the field F, i.e. we use the field’s multiplicative and additive identities rather than the integers 1 and 0.
With this more specific definition, we can write any matrix B ∈ Mm×n(F) with the following linear combination of elements of S.
in much the same manner as we did
before, as the set {Aℓk : Aℓk ∈ Mn×n
,k ∈{1,…,n},ℓ ∈{1,…,n},k ≤ ℓ}, however, we need to redefine
Aℓk to get the symmetry we desire. Putting it to δk(i)δℓ(j) + δk(j)δℓ(i) gets us our symmetry, but
has the annoying side effect of resulting in a value of one being added to itself whenever ℓ = k. Hence,
we need to “neglect” one of the terms whenever ℓ equals k. To do this, simply multiply one term by
(1 −δk(ℓ))1 to
get the following entry-wise definition.
With this we can see that when k = ℓ,
and when k≠ℓ,
so we confirm the symmetry of each matrix in S.
Since the number of elements of S is governed by the possible values for k and ℓ, then we look to those for the size of S. The value of k ranges from 1 to n and ℓ will always be greater than or equal to k, meaning that when k = 1,k = 2,…,k = n the are n,n − 1,…,1 values of ℓ. Summing these we get
This is the size of S, as needed.
Now, any B ∈ Symn can be written as a linear combination of the elements of S in the following manner.
be the set {Aℓk : Aℓk ∈ Mn×n
,k ∈{1,…,n},ℓ ∈{1,…,n},k < ℓ}, but define Aℓk, entry-wise, in the
following way.
We can see this results in skew-symmetry,
as well as for when i = j (the diagonals)
which is required in Skewn.
Given that k ranges over 1,…,n, k < ℓ, and ℓ ranges over k + 1,…,n then the size of S is the following.
Now, any B ∈ Skewn can be written as a linear combination of the elements of S in the following manner.
in problem 3.
Letting A1 and A2 be
respectively, we know that W1 = {A1B : B ∈ M2×2(F)} and W2 = {A2B : B ∈ M2×2(F)} are subspaces of M2×2(F). Elements of W1 have thus form
 | (6.1) |
and likewise elements of W2 are
 | (6.2) |
Since the identity matrix is in M2×2(F) then A1 ∈ W1 and A2 ∈ W2, and therefore they are both in W1 ∪W2. If this union were a subspace, then
would need to be in it, but this is neither in the form of equation 6.1 nor equation 6.2. So W1 ∪ W2 is not a subspace.
and therefore au + bv ∈ W1 + W2. Thus W1 + W2 is a subspace of V .
that is both symmetric and anti-symmetric must maintain the condition Aij = Aji = −Aji. Only the
zero matrix of Mn×n
does this, so Symn ∩ Skewn = {0}.
Let A ∈ Mn×n
and define matrices B and C by the following.
With these definitions, B,C ∈ Mn×n
. We then also have that
and
which indicate that B ∈ Symn and C ∈ Skewn. Looking at the sum of these two matrices
we see that it’s simply A, implying that A ∈ Symn ⊕ Skewn.
With the above two results, we have that Mn×n
= Symn ⊕ Skewn.
Now let h ∈ C0(ℝ, ℝ) and define the functions f and g by the following.
With these definitions, f,g ∈ C0(ℝ, ℝ). Given these we see that
and
which informs us that f ∈ 𝔈 and g ∈ 𝔒. Looking at the sum of f and g,
we see that it is simply h. So h ∈ 𝔒 ⊕ 𝔈.
By the results of the two paragraphs above, we have that C0(ℝ, ℝ) = 𝔒 ⊕ 𝔈
While there are some things in the solutions above that appear obvious to me, there are others that did not. For the latter, these are some of the proofs I generated in order to convince myself of certain properties.
Lemma A.1 (Commuting Scalars Among Matrices) Given a scalar c ∈ F, A ∈ Mm×ℓ(F), and B ∈ Mℓ×n(F) for some field F, c can commute about the matrix multiplication of A and B, that is
 | (A.3) |
Proof. In this case, it suffices to show that the ijth entry on the left-hand-side of A.3 is equivalent to the right-hand-side. Given this we have the following due to the multiplicative commutativity of F.
