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Bases
We start up with Linear independence:
The set {v1 , v2 ,
...
The alternate state of affairs, Linear
Dependence (LD) means that the vectors CAN be linearly combined to the zero vector
WITHOUT resorting to multiplying everything by zero
...

Theorem:
A set {v1 , v2 ,
...

Proof
...
Start with ONE of the vectors being expressible as a linear combination of the others
...

v1 = a2 v2 + a3 v3 + · · · + an v3
can be rewritten by subtracting v1 from each side for
0 = (−1)v1 + a2 v2 + · · · + an vn
which makes the set linearly dependent (since the coefficients are not all equal to zero)
...
We start with
0 = a1 v1 + a2 v2 + · · · + an vn
There’s at least one ak in there that’s not equal to zero
...
So, subtract an vn from each side for
−an vn = a1 v1 + a2 v2 + · · · + an−1 vn−1
and divide each side by −an for a final expression
vn = −

a1
a2
an−1
− v2 − · · · −
vn−1 ,
an an
an

and done
...
, vn } = Span{v2 , v3 ,
...

Proving this is not too hard, it’ll probably be on an assignment (maybe with a specific
number of vectors, or something of the sort)
...
If the set is not Linearly Independent, then it’s got unnecessary
terms (with respect to the span), which can be removed due to this result
...

Algorithm (to make a minimal spanning set):
1
...
, vn }
...
Get the full solution set to the problem
a1 v1 + a2 v2 + · · · + an vn = 0
3
...
e
...
Remove it from the set (so we get a new n value, etc)
...

4
...

So, that procedure (which is not fun to use, I can tell you) can take any FINITE SIZED
spanning set and break it down until it is linearly independent, and therefore the smallest
it can be while still spanning the set
...
Setting up the LI/LD problem leads to the equations
a=0

b + c + 2d = 0

a + 2b + c + 3d = 0
...

Add that to the second equation for b + d = 0
...

This DOES NOT make the fourth element the variable equation, it makes the second, third,
and fourth variable
...
We can remove the second to fourth, but the
fourth one is the most complicated, so take
     
0
0 
 1
 0 , 1 , 1 
...

Property: If a set {v1 , v2 ,
...
, vm , x}
is a linearly independent set
...

This one is fairly easy
...

This is not possible for b ̸= 0, so we get
a1 v1 + a2 v2 + · · · + am vm = 0,
which has only the trivial solution (the v are LI)
...

Definition: a set is maximal linearly independentin V if you can not include any additional
vector from V in the set without making it linearly dependent
...
Start with a linearly independent set {v1 , v2 ,
...

2
...
, vn }
...
If there is a vector x ∈ V that is NOT in W :
(a) include it in the set, so {v1 , v2 ,
...

(b) Go back to step 2
...
If there are no vectors in V outside W , then W = V and you are done
...
It is not, however, a spanning set for R3 , it actually represents the
plane −x − y + z = 0
...
How about we
just expand it to
     
0
1 
 1
 0 , 1 , 0 
...

Definition: A Basis for the vector space V is a linearly independent spanning set of V
...

Less Basic Examples:
     
0
1 
 1
 1  ,  1  ,  0  for R3


0
1
1

{x2 − 1, x2 + x, x − 1} for P2
...

Dimension
The Fundamental Theorem (of vector spaces?)
Take a vector space
V = Span{v1 , v2 , v3 ,
...
, xk } ∈ V
...
)
We always get k ≤ n
...
This one is a
proof by contradiction
...
So, we start by assuming that k = n + 1 then show that this makes the
set of x vectors linearly dependent
...

The first step is to write x1 as a linear combination of the v vectors:
x1 = a 1 v1 + a 2 v2 + · · · + a n vn
...
Also, x1 ̸= 0, since the set of
them is not linearly dependent, so at least ONE of the ak values is ̸= 0
...

a1
a1
a1
Since v1 is a linear combination of x1 and {v2 ,
...
, vn } = Span {x1 , v1 , v2 ,
...
, vn }
...

Again, we can be sure that there is some a value there that is ̸= 0
...
Reorder so that the a2 ̸= 0
...

a2
a2
a2
a2

We do the same thing yet again, leading eventually to
Span {v1 , v2 ,
...
, vn }
...
, vn } = Span {x1 , x2 , x3 ,
...

But wait, that means
xn+1 ∈ V = Span {x1 , x2 , vx ,
...

So, what does this mean? It puts a limit on the size of linearly independent sets in the
vectors space V equal to the size of the smallest spanning set
...
, xn } and {v1 , v2 ,
...

This is easy to prove: each set is both linearly independent and a spanning set for V , so
we apply the Fundamental Theorem both ways for n ≤ m and m ≤ n, so m = n
...
It’s a fundamental property of the vector space
...
, vn } (n vectors total) is any basis for vector space V , then the
dimension of V , Dim(V ), is equal to n
...

Notice that if we can find a linearly sized spanning set then we can get a basis using the
algorithm from earlier
...
One of these (only) is true:
5

1
...
Every independent set in V
has size ≤ n
...

2
...
Independent sets in V can be of ANY
size
...

Examples:
Here’s the basics:








n
R ,







Pn ,

1
0
0

...


...


...












,
...


...


{ n n−1 n−2
}
x , x , x ,
...


n + 1 of them, so Dim(Pn ) = n + 1
...
So, M2×2 has dimension 4
...

1
1 0

Planes of R3 that are subspaces are two dimensional, etc etc
...
How do we establish that? By proving that
ANY finite set is NOT a spanning set
...
, pn (x)}
...
Give
those the numbers
{Y1 , Y2 ,
...

Next, take the biggest one: Y = max{Y1 , Y2 ,
...
Now, is the polynomial xY +1 ∈ P
within Span{p1 , p2 ,
...
It’s not a spanning set
...

Now some equivalent properties:
Theorem:
The set {v1 , v2 ,
...
a basis of V
2
...
a maximal LI set in V
4
...
a spanning set for V , with Dim(V ) = n
Exercises:
Section 5
...
3:
4bd

7



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