Linear Algebra

A

A vector space is a nonempty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms (or rules) listed below. The axioms must hold for all vectors u, v, and w in V and for all scalars c and d.

The sum of u and v, denoted by u + v, is in V.

u + v = v + u

(u + v) + w = u + (v + w)

There is a zero vector 0 in V such that u + 0 = 0

For each u in V, there is a vector –u in V such that u + (-u) = 0

The scalar multiple of u by c, denoted by cu, is in V

c(u + v) = cu +cv

(c + d)u = cu + du

c(du) = (cd)u

1u = u

B

Axiom 1

Let (a,b),(c,d)∈R^2.

Then, (a,b)+(c,d)=(a+c,b+d)∈R^2.

Thus, Axiom 1 holds.

Axiom 2

Let (a,b),(c,d)∈R^2.

Then (a,b)+(c,d)=(a+c,b+d)=(c+a,d+b)=(c,d)+(a,b).

Thus, Axiom 2 holds.

Axiom 3

Let (a,b),(c,d),(e,f)∈R^2.

Then ((a,b)+(c,d))+(e,f)=(a+c,b+d)+(e,f)

=(a+(c+e),b+(d+f))

(a,b)+((c,d)+(e,f))

Thus, Axiom 3 holds.

Axiom 4

Claim is that (0, 0) will work.

Let (a,b)∈R^2.

Then (a,b)+(0,0)=(a+0,b+0)=(a,b).

Thus, Axiom 4 holds.

Axiom 5

Let (a,b)∈R^2. We need to find –(a,b). Recall that –(a,b) is the same as (–a,-b).

Claim is that (-a,-b) works.

(a,b)+(-a,-b)=(a+(-a),b+(-b))=(a-a,b-b)=(0,0)=0

Thus, Axiom 5 holds.

Axiom 6

Let m∈R, where R is the set of real numbers, and let (a,b)∈R^2.

Then, m(a,b)=(ma,mb)∈R^2.

Thus, Axiom 6 holds.

Axiom 7

Let m∈R, where R is the set of real numbers, and let (a,b),(c,d)∈R^2.

Then, m((a,b)+(c,d))=m(a+c,b+d)

=(m(a+c),m(b+d))

=(ma+mc,mb+md)

=(ma,mb)+(mc,md)

=m(a,b)+m(c,d)

Thus, Axiom 7 holds.

Axiom 8

Let m,n∈R, where R is the set of real numbers, and let (a,b)∈R^2.

Then (m+n)(a,b)=((m+n)a,(m+n)b)

=(ma+na,mb+nb)

=(ma,mb)+(na,nb)

=m(a,b)+n(a,b)

Thus, Axiom 8 holds.

Axiom 9

Let m,n∈R, where R is the set of real numbers, and let (a,b)∈R^2.

Then, m(n(a,b))=m(na,nb)

=(m(na),m(nb))...