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Subspaces:
The definition: W is a subspace of V if it is simultaneously a vector space with the same
addition and multiplication as V and if its vectors are a subset of V ’s vectors
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They are typically written W ⊆ V
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Those are the ‘trivial’ subspaces
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Anything
else is classified as a ‘proper’ subspace
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Example: First: you Can Not describe R2 as a subspace of R3 , since their vectors are
completely incompatible, as are their addition and multiplication
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Example: P2 ⊆ P3
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Example: The space of polynomials P is a subspace of the space of functions F
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We can confirm this with
 
 
4
2
 0  is on the plane
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The vector v =
0
0
NOT, as it does not satisfy the equation
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The closure axiom
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All possible subspace candidates need to use the SAME ADDITION AND MULTIPLICATION
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The Subspace Test:
There are three things that must be checked to confirm that a set of vectors W within
V is a subspace
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Theorem: Assume that all vectors in W are in V
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The zero vector (0 ∈ V ) is in W
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W is closed under vector addition
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W is closed under scalar multiplication
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We will now prove that W must then have ALL the
necessary properties of a vector space
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Now to go
through the ten axioms
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A2 and A3 are true as the addition is the same as for V
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A5 is done via property 3: simply use −1 × x (or (−1) x, if you prefer)
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S1 is property 3
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That’s all of them,
the proof is now finished
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Now, some may notice that the first requirement could be more general, simply
a ‘W is non-empty
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This is true, but the zero test should be there for a few
reasons
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• Finally, not having 0 in W is very common in cases where W is not a subspace, so it
is a good strategy to check that first
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Example: Homogeneous equations are of the form an xn + an−1 xn−1 + · · · + a1 x + a0 = 0, a
zero constant term
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The zero element in E3 is simply 0x3 + 0x2 + 0x1 = 0
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Adding:
a3 x3 +a2 x2 +a1 x1 = 0

b3 x3 +b2 x2 +b1 x1 = 0

=⇒

(a3 +b3 )x3 +(a2 +b2 )x2 +(a1 +b1 )x1 = 0

so yes
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Example: Is V = {f (x)|f (1) = 0} a subspace of F(0,2) ?
We just need to check the three parts of the subspace test
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Now to check addition
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(g + h)(x) = g(x) + h(x),

(g + h)(1) = g(1) + h(1) = 0

so V is closed under addition
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ag(1) = a(0) = 0
so yes, it is a subspace
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, vn } ∈ V is the set of
all linear combinations of {v1 , v2 ,
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So:
Span{v1 , v2 ,
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, an ∈ R}
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In essence, everything you can make out of
the vectors v1 to vn , using vector addition and scalar multiplication
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First: the zero vector:
0 = 0v1 + 0v2 + · · · + 0vn ,
which is a perfectly reasonable linear combination
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Take two elements
a = a1 v1 + a2 v2 + · · · + an vn
and
b = b1 v1 + b2 v2 + · · · + bn vn
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Next we check multiplication
c a = c (a1 v1 + a2 v2 + · · · + an vn )
= (ca1 )v1 + (ca2 )v2 + · · · + (can )vn ,
yet another linear combination
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, vn }, all vk ∈ V , is a Subspace of V
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Namely:

3

If v1 , v2 ,
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, vn }
is a subspace of W
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We can expand this even further
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, vn } is the smallest possible subspace containing {v1 , v2 ,
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How do we get this? It’s actually quite easy
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, vn } and that all v vectors belong to W
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, vn } ⊆ W , which combines with the other ⊆ to give us
Span{v1 ,
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So, the span of a set of vectors in V is always the smallest subspace of V containing that
set of vectors
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It also fits in nicely with
one description of spans: they are everything you can make out of a set of vectors using the
vector space operations
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Example:
Lets look at the span of the following set
{ 2
x + x, x2 + 1,

}
1 ∈ P2
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Four variables, three equations
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This makes
a2 − b2 + b3 = a1

=⇒ quadb3 = a1 − a2 + b2
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We’ve satisfied every equation, and here’s the result:
b1 = a2 − b2

b3 = a1 − a2 + b2

b4 = a0 + a1 − a2 − b2

with b2 arbitrary
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That set of vectors actually spans the whole space
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4

Spanning Sets:
We will spend some time finding these
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You have a vector space V , and a set {v1 , v2 , v3 ,
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, vn }
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Efficiency is
NOT part of the definition
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The previous example spanned P2 , but the simpler one is
{
}
1, x, x2 , ,
however, this works too:

{
}
1, x, x2 , 0, 12, −x2 + 1, x2 + 2
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Here’s another basic example:
     
0
0 
 1
 0 , 1 , 0 
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Spans create vector spaces, so
spanning sets only really apply to vector spaces
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So, for instance,
any plane in R3 of the form
x = bt + cs

t, s ∈ R

is a subspace, since that is a span, etc
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• Spanning sets allow you to characterize an entire vector space using only a small set
of vectors
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• Remember: spanning sets need to be INSIDE the space in question and span it
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They can have unnecessary terms
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1
6
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deg(p(x)) = the highest power of x in the polynomial
7
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b)d)f)
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b)d)

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