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Week of...
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Notes and Links
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1
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Sep 7
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Tue, About, Thu
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2
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Sep 14
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Tue, HW1, HW1 Solution, Thu
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3
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Sep 21
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Tue, HW2, HW2 Solution, Thu, Photo
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Sep 28
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Tue, HW3, HW3 Solution, Thu
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Tue, HW4, HW4 Solution, Thu,
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Tue, Thu
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Oct 26
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Tue, Why LinAlg?, HW6, HW6 Solution, Thu
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Tue, MIT LinAlg, Thu
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Tue, HW9, HW9 Solution, Thu
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13
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Nov 30
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Tue, On the final, Thu
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Office Hours
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Dec 14
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Final on Dec 16
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The Algebra Song!
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Misplaced Material
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Replacement Theorem
...
-
dim(V) = n
- If G generates V then
. If also
then G is a basis.
- If L is linearly dependent then
. If also
then L is a basis. If also
then L can be extended to a basis.
Proofs
a1. If G has a subset which is a basis then that subset has n elements, so
.
a2. Let
be a basis of V, then
. Now use replacement with G & L =
. Hence,
.
a. From a1 and a2, we know
. If
then G contains a basis
. But
, so
, and hence G is a basis.
b. Use replacement with G' being some basis of V. |G| = n.
If
then
, so
, so
generates, so L generates, so L is a basis since it is linearly independent.
We have that L is basis. If
the nagain find
such that
and
generates.
1.
generates V.
2.
. So by part a,
is a basis.
3.
-
If V is finite-dimensional (f.d.) and
is a subspace of V, then W is also finite, and
. If also dim(W) = dim(V) then W = V and if dim(W) < dim(V) then any basis of W can be extended to a basis of V.
Proof: Assuming W is finite-dimensional, pick a basis
of W;
is linearly independent in V so by Corollary 3 of part b,
. So span(
) = V = W so V = W. ...
Assume W is not finite-dimensional.
so pick a
such that
. So
is linearly independent in W, and
. Pick
. So
is linearly dependent and
. Pick
... continue in this way to get a sequence
where n = dim(V) and
is linearly independent. There is a contradiction by Corollary 3.b.
The Lagrange Interpolation Formula
[Aside:
because
]
Where
Let
be distinct points in
.
Let
be any points in
.
Can you find a polynomial
such that
? Is it unique?
Example:
![{\displaystyle x_{i}=0,1,3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6421a9ff670159252be5c6529b17155be379f664)
![{\displaystyle y_{i}=5,2,2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/818604bbf341e5503317b62566a560265d7223e9)
Can we find a
such that
![{\displaystyle P(0)=5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65f18660f71352890a3f170867a14b60ff85dc08)
![{\displaystyle P(1)=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f01ea5fa74d85fb9469585114c1f662e9feeda13)
?
Solution: Let
. (Remember capital pi notation.)
Then
![{\displaystyle {\tilde {P}}_{1}=(x-x_{2})(x-x_{3})=(x-1)(x-3)=x^{2}-4x+3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7075b159c218e6ea05056243f1d9f7bcc4a32a5d)
![{\displaystyle {\tilde {P}}_{2}=(x-0)(x-3)=x^{2}-3x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e744f660dea17445d5bac2ddec9da33d10a57fa1)
![{\displaystyle {\tilde {P}}_{3}=(x-0)(x-1)=x^{2}-x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69f3dd5a657a0ceff4f742bd1c8e30bfae1b897f)
Set
Then
![{\displaystyle P_{1}(x)={\frac {1}{3}}x^{2}-{\frac {4}{3}}x+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71eb54f9789e62f9c2b00a9820947b5adb42fd7e)
![{\displaystyle P_{2}(x)=-{\frac {1}{2}}x^{2}+{\frac {3}{2}}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ea2b71470295abd058cc2832755f7cb397b7b6)
![{\displaystyle P_{3}(x)={\frac {1}{6}}x^{2}-{\frac {1}{6}}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06c45551234888248a3756956300822980294d4b)
Let
![{\displaystyle P=5\cdot \left({\frac {1}{3}}x^{2}-{\frac {4}{3}}x+1\right)+2\cdot \left(-{\frac {1}{2}}x^{2}+{\frac {3}{2}}x\right)+2\cdot \left({\frac {1}{6}}x^{2}-{\frac {1}{6}}x\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03d46a2cb372f5c1f79f84ec315ef7bbb689bfa5)
![{\displaystyle \,\!=x^{2}-4x+5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce272e3a32a6f7845a22bbe6b374e9125c517d94)