Exercise 1 Let Mn (F) denote the space of n x n matrices with ent...

Exercise 1
Let Mn (F) denote the space of n x n matrices with entries in the field F.
(a) Let Sn(R) denote the set of symmetric matrices in Mn (R). Show that Sn (R) is a subspace of Mn (R). (A square matrix is symmetric if it is equal to its transpose).
(b) Let Hn(C) denote the set of Hermitian matrices in Mn (C). Show that Hn(C) is a subspace of Mn (C).
(A square complex matrix is Hermitian if it is equal to its Hermitian conjugate or Hermitian trans-pose. The Hermitian conjugate of a matrix A is obtained by transposing A and then replacing every entry with its complex conjugate).

Exercise 2
Show that the set c of convergent sequences of real numbers is a vector space over R. (To do this you need to show that the sum of two convergent sequences is convergent, and that any scalar multiple of a convergent sequence is convergent).

Definition 2 (Loose definiton: linear combination, linear span, spanning set)
Suppose that S is a subset of a F-vector spaces V. An element v of V is a F-linear combination of elements of S if v = a1 s1 +...+ ak sk,
where k is a positive integer, and s1, ... ,sk are elements of S, and a1, ..., ak are scalars. The set of all linear combinations of elements of S in V is called the linear span of S and denoted by (S) or span(S). If S = {ul,u2,...,ut} is a finite set, we can denote the span of S by (ui, u2, .... ,ut). If every element of V is a linear combination of elements of S, then S is called a spanning set for V.

Exercise 3
If S is a subset of a F-vector space V, prove that (S) is a subspace of V. (This means showing that (S) is closed under addition and scalar multiplication).

Exercise 4 In Q[x], show that x3 + x2 + x + 1 is not a linear combination of x3 + x2, x2 + x and x + 1.

Definition 3 (Linear independence)
Let V be a IF-vector space. A subset S of V is linearly independent (over F) if no element of S is a IF-linear combination of other elements of S.

Exercise 5 Suppose that S is a subset of a F-vector space V. Suppose there exist distinct elements s1, s2,..,sk of S for which
s1 =Σᵢ₋₂ᵏ ai si, ai ∈ F for i = 2,..., k
Prove that〈S〉=〈S\{s1}〉

Exercise 6 Give an example of a linearly independent subset of Q[x] with five elements, and an example of a linearly dependent subset of Q[x] with five elements.

Exercise 7 Let S be a subset of a F-vector space V. Prove that S is linearly independent if and only if every element of 〈S〉can be expressed as a linear combination of the elements of S in a unique way.

Definition 4 (Basis)
A subset B of a vector space V is a basis of V if and only if every element of V can be expressed in a unique way as a linear combination of the elements of B.
Thus a basis of a vector space is a spanning set that is linearly independent. A vector space is called finite—dimensional if it has a finite spanning set. Otherwise it is called infinite-dimensional. We will mostly be interested in finite dimensional spaces in this course.

Exercise 8 Give two examples of finite-dimensional vector spaces and two examples of infinite-dimensional vector spaces.

Exercise 9 Suppose that V is a finite-dimensional F-vector space and that S is a finite spanning set for V. Show that some subset of S is a basis of V.
The last part of this note and set of exercises is to show that if V is a finite-dimensional vector space, then every basis of V has the same number of elements, this number is called the dimension of V.

Exercise 10 Let V be a finite-dimensional vector space over a field F, and suppose that V has a basis B with n elements.
(a) Show that a linearly independent subset of V can have at most n elements.
(b) Show that a spanning set for V must have at least n elements.
(c) Conclude that every basis of V must have exactly n elements.

A key strategy for Exercise 10 is the Steinitz replacement lemma, the basic mechanism works as follows. For part (a), suppose that S = {s1,...sk} is a linearly independent subset of V. We need to show that k can't exceed n.
The element Si of S can be written in a unique way as a linear combination of elements of B (why?). Let b1 be an element of B that appears with non-zero coefficient in this expression. Then B1 := B \{b1} U {s1} is another basis of V (why?). Now the expression for s2 as a linear combination of the elements of B1 can't involve only s1, it must involve some element b2 of the original B (why?). Then we can replace b2 with s2 to get a new basis B2 of V. What happens when we get to sn?
The same idea can be adapted for part (b). Although you have probably seen these arguments before, it's worth recalling them in detail at the start of this course.
If B = {b1,...,bn} is a basis for a F-vector space V of dimension n, then every element v of V can be expressed in a unique way as a linear combination of b1,...,bn, with coefficients which are elements of F. These coefficients can be thought of as the coordinates of v, with respect to the basis B. This means that every element of V can be represented as a column or row vector with entries in F. Thus V can be identified (in many different ways depending on the choice of basis), with the space IF" of column vectors of length n over F.