Lecture 12

A. Agarwal
August 6, 2012

Recall

  1. 1.

    Vector space: (V,+,) with ten axioms

  2. 2.

    Subspace: a subset of a vector space that is a vector space on its own.

Recall that to ensure that a subset SV is a subspace of V, we must have that

  1. 1.

    S is nonempty.

  2. 2.

    S is closed under addition.

  3. 3.

    S is closed under scalar multiplication.

These last two items can be understood as S being closed under linear combinations. That is, S is a subspace if for every u,vS and c1,c2 (scalars), then c1u+c2vS.

Defining Operations

The usual definitions for addition and multiplication are not the only ones possible. Let us look at some examples. Consider the set

V={(xy):x,y}

with addition and scalar multiplication defined as

(xy)+(ab) =(x+ay+b)
c(xy) =(xcy)

Question:

Is V with these operations a vector space or not?

To answer this, we must check each of the requirements of a vector space.

Additive closure

Since this is the usual definition of addition, it is closed under addition.

Scalar multiplication

To check whether the scalar multiplication defined here satisfies the axioms, let (xy)V and c. Then c(xy)=(xcy) and since x,cy, then (xcy)V, hence V is closed under scalar multiplication.

Zero vector

Claim 1.

(00)V is the zero vector.

Proof.
(xy)+(00)=(x+0y+0)=(xy)

Since addition here is the same as the usual addition, if there is a problem with these definitions that prevents the space from being a vector space, it must have something to do with the multiplication operation.

Distributivity

A vector space must satisfy distributivity: (c+d)u=cu+du

To check whether this holds in this space, let u=(xy) and c,d. Then

(c+d)u=(c+d)(xy)=(x(c+d)y)
cu+du=c(xy)+d(xy)=(xcy)+(xdy)=(2x(c+d)y)

Since x will not always equal 2x, the result fails and we can conclude that V with these operations is not a vector space.

Though this set with these operations is not a vector space, can it be that perhaps other operations would maintain the vector space structure?

Consider the set

V={x|x>0}=+

For clarity, define addition and scalar multiplication as follows:

xy =xy
cx =xc

where x,y+ (the vector space) and c (a scalar).

Question:

Is (+,,) a vector space?

  1. 1.

    Closure under

    Let x,yV (so x,y>0). Then

    xy =xy>0

    Therefore xyV.

  2. 2.

    Closure under

    Let c and xV. Then since x>0,

    cx=xc>0

    Therefore cxV.

  3. 3.

    Zero element

    Suppose some element zV is the zero element. Then for any xV,

    xz =x
    xz =x
    x(z-1) =0

    Thus z=1 would have to be the zero element. Since 1>0, thus 1V and so the zero element is in fact z=1.

  4. 4.

    Inverse of an element

    Let xV. Then we want some tV such that

    xt =tx=z

    where z is the zero vector. Remember that we have just found that z=1 is the zero element, rather than the usual z=0. Carrying out these operations, we have that

    xt =1
    t =1x

    Since xV, thus x>0, so 1x>0. Thus t=1xV. That is, for each xV, the element 1x is its additive inverse.

  5. 5.

    To check the axiom:

    (c+d)u=cudu

    where c,d and u=xV (so x>0).

    Starting with the left-hand side:

    (c+d)u =(c+d)x
    =xc+d
    =(xc)(xd)
    =(cx)(dx)
    =(cx)(dx)
    =cudu

    we arrive at the right-hand side, indicating the axiom holds.

It turns out that this set does in fact satisfy all of the vector space axioms. With such strange operations, it brings to light that our standard vector spaces are by no means the only ones.

Question:

Is (V,+,)(V,,)?

Subspaces of 2

You may recall that all the subspaces of 2 are:

  1. 1.

    2

  2. 2.

    {(00)}

  3. 3.

    All straight lines through the origin.

To see why this is the case, let us outline the proof:

Proof.

Let S be a subspace of 2. Note that if S={(00)}, then S is indeed a subspace. Otherwise, let uS such that u0. Then since S is a subspace, thus cuS for all c. Thus it contains the straight line along u that passes through the origin.

If S has no other vector, then S is simply that straight line. Otherwise, assume there exists some non-zero vS that is linearly independent of u.

Note then that span{u,v}=2 (Why?) ∎

The above proof is not quite complete. The question is why we can gaurantee that span{u,v}=2?

What this amounts to is whether for any vector w2 there exist some scalars c1,c2 such that

c1u+c2v =w

This is a linear system with corresponding augmented matrix looking like:

[uvw]

Since ucv, that is, they are linearly independent, then the coefficient matrix portion of this augmented matrix is invertible. Thus its columns will span 2.