OVERVIEW:

A vector space is like a family or private club for vectors whose members relate to each other according to 2 well defined operations and a set of rules of behaviour known as axioms. Because we're used to working with vector spaces like the number line and the Cartesian Plane, we take these behaviour axioms for granted. However, with vectors and their spaces, we sometimes re-define addition and scalar multiplication in strange ways, in which case we must test all the axioms for validity under these newly defined operations.

The "vectors" or elements in a vector space are not necessarily what we generally call vectors. A single point, a set of numbers, matrices, polynomials, lines, or planes are all elements of vector spaces, because we can show that the 10 axioms apply to specifically defined groups of all these objects.

Once a set of vectors has been identified as a vector space or "private club", any subset can be called a subspace if we can show that this subset satisfies just 2 of the 10 axioms.

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VECTOR SPACES DEFINED

Def'n: A group of objects upon which addition and scalar multiplication have been defined is called a vector space, if the following 10 axioms are satisfied:

1) if , then . (closure under addition)

2) u + v = v + u (addition is commutative)

3) u + (v + w) = (u + v) + w (addition is associative)

4) there is a 0 in V such that 0 + u = u + 0 = u (additive identity)

5) for all , there exists (– u) such that u + (– u) = 0. (additive inverse)

6) for any real number k, . (closure under real number scalar multiplication)

7) k(u + v) = ku + kv. (scalar on a vector sum distributive property)

8) (k + l)u = ku + lu. (scalar sum on a vector distributive property)

9) k(lu) = (kl)u. (commutative property under scalar multiplication)

10) 1u = u. (multiplicative identity = 1, a scalar)

Notes:

axioms 1, 2 & 3 say: the sum of 2 vectors is in V, the order & grouping of the sum don't matter.

axiom 4 says: there's a zero vector 0 -- that changes nothing when added

axiom 5 says: there's an inverse vector ( – u) -- that added to u takes us back to 0.

axiom 6 says: any Real scalar multiple of a vector in V, is also in V.

axioms 7, 8 & 9 say: scalar multiplication of vectors is distributive and commutative.

axiom 10 says: when we multiply by 1 (the scalar), it changes nothing.

The Zero Vector is an additive identity -- and is not always made up of only zeros. If either operation, addition or scalar multiplication, is defined in a non-standard way; if the zero vector exists, it must satisfy the condition that 0 + u = u + 0 = u -- add it to another vector and change nothing! If no such vector exists in the set, it has no 0 and is not a vector space.

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Axiom 10 mentions the scalar "1" -- the multiplicative identity element -- but unlike the zero vector, whose components can be varied to satisfy the required condition -- here we're talking about the number one -- which is immutable -- so we must pay close attention to verifying the axioms that involve scalar multiplication.

When one (or both) of the operations -- addition & scalar multiplication -- is defined in a non-standard way, we test for validity all the axioms mentioning that operation.

Example: determine if the set of all triples (x, y, z) with standard addition

(x, y, z) + (x', y', z') = (x + x', y + y', z + z')

but scalar multiplication defined by k (x, y, z) = (kx, y, z), forms a vector space.

This new definition of scalar multiplication instructs us to multiply only the 1st entry by the scalar k -- instead of the standard way -- where we multiply all 3 entries by k.
This tells us that we must verify all the axioms concerned with scalar multiplication.

We find that the set satisfies axioms 1 through 7; but won't satisfy axiom #8 because:
( kx, y, z ) + ( lx, y, z ) = ( (k + l)x, 2y, 2z ) instead of ( (k + l)x, y, z )

Now say the question defined scalar multiplication as k (x, y, z) = (2kx, y, z), axiom 10 would fail as well because 1(x, y, z) would now = (2x, y, z) which is not the same vector as (x, y, z).

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Under the standard definitions of addition and scalar multiplication, a line or plane through the origin is a Subspace of R³. The origin (0, 0) itself forms a vector space in R³ since it satisfies all axioms.

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Example:

Determine whether the set of all vectors in R² in the first and third quadrants forms a vector space. If it doesn't, state which axioms fail.

All 2-dimensional vectors in the 1st and 3rd quadrant can be represented by

.

Axiom 1 says that the sum of any 2 vectors in the set must also be in the set.

If we consider the vectors , their sum is
which is in the 2nd quadrant.

Axiom 1 fails the set is not closed under addition, so it is not a vector space.
All other axioms are satisfied.

Note: a single counter-example is enough to DISPROVE a concept or property, however,
any number of examples is not enough to prove anything. To show an axiom fails, we need only show one instance where it does (as I did above).

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Example:

Show that all 2-dimensional vectors on the x-axis form a vector space.

The 2-dimensional vectors on the x-axis can all be generated from multiples of ,
since they all have y = 0. All the axioms will be satisfied under the standard definitions of addition and scalar multiplication since 2-dimensional vectors behave like Real Numbers under those operations.

The Axiom proofs will look like this:

Axiom 1: closed under addition -- sum is still on the x-axis (y = 0)

Axiom 2: summing order doesn't matter.

Axiom 6: scalar product is still on the x-axis (y = 0).

and so on . . . use (x, y, z) with (x', y', z') or (u₁, u₂, u₃) with (v₁, v₂, v₃) to represent the vectors.

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SUBSPACES

to determine if W, a subset of vectors from a vector space V, forms a subspace of V,
we need only verify axioms 1 and 6 -- for closure under addition & scalar multiplication.

If W is closed under addition and scalar multiplication, it is a subspace of V.

Example: Determine if the set of all vectors on the xy-plane form a subspace of R³ .

Any vector on the xy-plane has z = 0 and looks like this:

Since the sum of 2 such vectors is still on the xy-plane with z = 0
and any scalar product of such a vector will still have z = 0, the set is
closed under addition and scalar multiplication -- so it forms a Subspace of R³ .

This is a 2-dimensional subspace of R³ -- a plane through the origin.

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Thm: The solution set of a homogeneous system of linear equations Ax = 0 is a subspace of Rⁿ.

Proof: if x and x₁ are elements of the solution set of Ax = 0, then
A(x + x₁) = Ax + Ax₁ = 0 + 0 = 0 (closed under addition)
and A(k x) = k(Ax) = k(0) = 0 (closed under scalar multiplication)
Therefore the solution set of Ax = 0 is a subspace of Rⁿ.

FOR MORE INFO, EXAMPLES AND PRACTICE ON THIS TOPIC --
see lessons lnalg 4, 4.1, 4.2 , assignments #4 and #5, as well as Test #3

( Linear Algebra MathRoom Index )

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