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More on Spans
Spans are a frequently difficult concept
...

Definition:
Span {v1 , v2 , v3 ,
...
, an ∈ R}

so ALL the linear combinations of the vectors in the set
...

• The span of {v1 , v2 ,
...
, vn
...

As in, we’ve seen examples like
{
}
W = a2 x2 + a0 , a2 , a0 ∈ R ,
which is a subspace of P2
...

Having it writable as a span (EXACTLY writable as a span) proves it to be a subspace
...


Write it as a span and you have a subspace, vector space, etc
...
Look at the form of those vectors:
(a3 − a2 + a1 )x2 + (2a2 − 3a1 )x + 4a3
= a3 x2 − a2 x2 + a1 x2 + 2a2 x − 3a1 x + 4a3
= a3 (x2 + 4) + a2 (−x2 + x) + a1 (x2 − 3x),

a1 , a2 , a3 ∈ R,

a span, so it’s a subspace of P2
...
A little easier than checking the properties, though
there are cases where that is necessary
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
 
  
−1
0 
 1
 −1  ,  2  ,  1 


2
−1
1

Does the set

span R3 , is it a SPANNING SET for R3 ? That means the set has to have ALL VECTORS
IN R3 in its span, its set of linear combinations
...


Later we’ll be able to be COMPLETELY certain about this sort of thing, but this basically
shows us that we get an answer only if 3x + y − z = 0, only on that plane
...

2
−1
1

This is a big deal
...
We need THREE DIFFERENT
directions, we really only have two
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and dependence, as well
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Consider the following
spanning set for R2 :
{[ ] [ ] [ ]}
1
1
2
,
,

...

Look at the last two
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As a result, a span of the two of them has two directions to work with and, as a result,
covers a plane, in this case R2
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1
1
1
1
Problem is, that’s a property of just two vectors
...
Notice, however, that there’s one commonality
between those two ̸= sets:
[ ]
[ ]
1
2
a
=b
means that a = b = 0
1
1
which is the exact same as
[ ]
[ ]
2
1
=0
+b
a
1
1

means thata = b = 0
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Now to bring back the full set of three
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The vectors going in different directions could only be linearly combined
to 0 if they were all multiplied by zero
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It’s not what you can make out of the vectors,
it’s how you can make it
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, vn } is Linearly Independent if
a1 v1 + a2 v2 + · · · + an v3 = 0

implies that a1 = a2 = · · · = an = 0
...
Note that this linear combination is ALWAYS POSSIBLE
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3

Definition: a set of vectors {v1 , v2 ,
...
Alternative Definition: if the set of vectors
is not Linearly Independent
...
It’s just that
linearly dependent sets have other options
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If v1 can be created out of the others:
v1 = b 2 v2 + b 3 v3 + · · · + b n vn
which converts easily to
0 = (−1)v1 + b2 v2 + b3 v3 + · · · + bn vn ,
so linearly dependent
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If {v1 , v2 ,
...

Example: Here’s two sets to consider linear dependence, etc,
{
}
{
}
A : x2 + x, x2 + 1, x + 1
B : x2 − x, x2 + 1, x + 1
A: need to check the available answers for:
a1 (x2 + x) + a2 (x2 + 1) + a3 (x + 1) = 0
which converts to
(a1 + a2 )x2 + (a1 + a3 )x + (a2 + a3 ) = 0,
a1 + a2 = 0,

a3 + a1 = 0,

a2 + a3 = 0

a1 = −a2 ,

a3 = −a1 = a2

a3 = −a2
...
ONLY THE ZERO ANSWER
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B: need to check the available answers for:
a1 (x2 − x) + a2 (x2 + 1) + a3 (x + 1) = 0
which converts to
(a1 + a2 )x2 + (−a1 + a3 )x + (a2 + a3 ) = 0
a1 + a2 = 0,

a3 − a1 = 0,

a2 + a3 = 0

a1 = −a2 ,

a3 = a1 = −a2 ,

a3 = −a2

This leads to
so we get multiple options, we can’t break it down until everything is zero
...

As a result, we have ourselves a Linearly Dependent set, LD
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• Any set of vectors that includes 0 is linearly dependent in a vector space
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These are fairly easy to show
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Not has hard as it looks
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a cos(x) + b sin(x) = 0,

Look at two points of x, x = 0 and x = π
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Example:

{[

3 2
−1 0

] [
] [
]}
1 −1
2 1
,
,
2 1
0 −1

Again, check every single solution of the system
]
]
]
] [
[
[
[
0 0
3 2
1 −1
2 1

...


Notice the last one, gives us a2 = a3
...


Using the first one we get a1 = −a2 , using the second a1 = 0, the last a1 = −2a2 , meaning
the only solution a1 = a2 = a3 = 0
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1
5
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b)d)

Section 4
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a)b)c),

2
...
2
1
...
b)d)f)h)
5



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