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Vector Space

e.g., \(\mathbb{R}, \mathbb{R}^2, \mathbb{R}^3, \mathbb{R}^n, \mathbb{C}, \mathbb{F}^n \text{(} F \text{ is a field)}, M_{m\times n} F, f(Set, Field), P(F)\)

Definition

A vector space over a Field \(F\) consists of a set on which two operators (addition and scalar multiplication) are defined, so that \(\forall x,y\in V, \exists\) unique element \(x+y\in V\) and \(\forall a\in F, x\in V, \exists\) unique element \(ax\in V\) so that the following conditions hold:

  1. \(\forall x,y\in V,x+y=y+x\)
  2. \(\forall x,y,z\in V, (x+y)+z=x+(y+z)\)
  3. \(\exists 0\in V,\forall x\in V, x+0=x\)
  4. \(\forall x\in V, \exists y\in V, x+y=0\)
  5. \(\forall x\in V, 1\cdot x=x\)
  6. \(\forall a,b\in F, x\in V, (ab)x=a(bx)\)
  7. \(\forall a \in F, x,y\in V, a(x+y)=ax+ay\)
  8. \(\forall a,b\in F, x\in V, (a+b)x=ax+bx\)

Theorem 1

Let \(x,y,z\in V\) st \(x+z=y+z\), then \(x=y\)

Proof

\(x=x+0\)

\(\exists v \in V, z+v=0\)

\(x=x+(z+v)=(x+z)+v=(y+z)+v=y+(z+v)=y+0=y\)

Extensions

Prove 0 is unique:

Suppose 0 is not unique, there must exist \(z_1\), \(z_2\) st \(z_1\neq z_2\) (zeros).

Thus, \(z_1=z_1+z_2=z_2\)

Thus, 0 is unique

Prove \(y\) is unique:

Suppose there is more than one y, denoted as \(y_1\) and \(y_2\) st \(y_1\neq y_2\).

Thus, \(y_1+0=y_2+0\), since there is only a unique zero, \(y_1=y_2\)

Thus, \(y\) must be unique

Prove additive inverse is unique:

Suppose there is more than one \(y\), denoted as \(y_1\) and \(y_2\) st \(y_1\neq y_2\).

\(x+y_1=0,x+y_2=0\)

Thus, \(y_1+x+y_1=y_1\), \(y_2+x+y_1=y_1\)

Thus, \(0+y_1=y_1, 0+y_2=y_1\)

Thus, \(y_2=y_1\)

Prove \(0x=0\)

Suppose \(0x=y, y\neq 0\)

\(2y=0x+0x=(0+0)x=0x=y\)

Thus \(2y=y, y=0\)

Vector SubSpace

Definition

A subset \(W\) of a vector space \(V\) over a field \(F\) is called a subspace if \(W\) us a vector space of \(F\) with operations of

addition and scalar multiplication defined on \(V\)

Ex

an \(\mathbb{R}^2\) in \(\mathbb{R}^3\)

Theorem 2

Let \(W\subset V\), \(W\) is a subspace of \(V\iff\) the following conditions hold.

  1. \(0\in W\)
  2. \(\forall x,y\in W, x+y\in W\)
  3. \(\forall c\in F, \forall x\in W, cx\in W\)

Partial Prove

The proof of any subset \(W\) with these properties contains the additive inverse.

Since \(F\) is a set and \(1\in F, -1\in F\)

Thus, \(\forall x\in F, -1*x\in W\)

Thus, \((1*x)+(-1*x)\in W\)

Thus, \((1-1)x=0x=0\)

Thus, the additive inverse exists in \(W\)