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Title: Linear Algebra - Bases (Part 2)
Description: These notes cover the second part of Bases
Description: These notes cover the second part of Bases
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Finishing Bases:
Definition: A minimal spanning set of V is a set that spans V but does not span V if a
single vector is removed
...
Definition: A maximal linearly independent set in V is a set that is LI, but becomes LD if
a single additional vector in V is included
...
Definition: A basis for vector space V is a spanning set for V that is also a linearly independent set
...
Turns out these are all the same thing
...
, vn } ∈ V is either ALL of the following or NONE:
1
...
a minimal spanning set of V
3
...
an independent set in V , where Dim(V ) = n
5
...
Example: Is the following space equal to R3 ?
0
2
1
Span 0 , 2 , 1
...
Painful
...
That’s not easy, but it’s not that hard
...
In a three dimensional space any three
linearly independent vectors form a basis, so done, they’re a spanning set
...
Proposition: If U is a subspace of V then Dim(U ) ≤ Dim(V )
...
This one is actually fairly simple
...
Proposition: If U is a subspace of V and Dim(U ) = Dim(V ) then U = V
...
Also fairly simple
...
It becomes a maximal independent
set in V and so is a basis
...
, vn } be a basis for vector space V
...
, an such that
x = a1 v1 + a2 v2 + · · · + an vn
...
First, we can get some values a since the vectors span V
...
Assume we have another set of values:
x = b1 v1 + b2 v2 + · · · + bn vn
so
x − x = 0 = (a1 − b1 )v1 + (a2 − b2 )v2 + · · · + (an − bn )vn
which is a linear combination for the zero vector, and the set of vectors is linearly independent
...
The expansions
are unique
...
0
0
0
1
Next, we’ll use
We need
2
1
0
1 , 1 , 1
...
2
2
0
0
The last line shows a = −b, which combines with the second line for c = 3
...
Final answer:
1
0
2
1
3 = 1 1 + (−1) 1 + 3 1
...
If {x, y, z} form a basis of vector space V then: does the set {x + y, y + z, z + x} form
a basis of V as well?
Not as hard as it looks
...
First: V is a vector space which has a 3 sized basis
...
As a result, a set, like the one we’re given, spans V if and only if it is
linearly independent
...
a(x + y) + b(y + z) + c(z + x) = 0
(a + c)x + (a + b)y + (b + c)z = 0
...
Of three Linearly Independent vectors
...
So:
a+c=0
a+b=0
b + c = 0
...
Using those on the last equation gives us −2a = 0 so a = 0
then a = b = c = 0
...
Now the same for {x − y, y − z, z − y}
...
We get c = a and b = a
...
So, we get a = b = c =free
...
Reducing a Spanning Set:
We do not want to do this too often, but lets take a look:
1
0
1
0
1
2 , 2 , 1 , 1 , 0
−1
1
0
−1
1
is a spanning set of R3
...
This may take a while
...
a = −c − f
...
Substitute that into the middle equation for
a = −c − f
− 3c + 3d − 6f = 0
b = −c + d − 2f
...
So, we can get rid of the vectors corresponding to a, c, d, f , NOT b, since it came out to zero,
it is NOT variable by this solution (which happens to be correct)
...
1
0
1
1
2 , 2 , 1 , 0
−1
1
0
1
4
and then look into
1
0
1
1
0
2 + b 2 + c 1 + d 0 = 0
...
Anyway, get rid of a vector associated with a, c, d and you’re done
...
−1
1
0
For exercises, see the previous section
...
5
Title: Linear Algebra - Bases (Part 2)
Description: These notes cover the second part of Bases
Description: These notes cover the second part of Bases