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1

Sep 7

Tue, About, Thu

2

Sep 14

Tue, HW1, HW1 Solution, Thu

3

Sep 21

Tue, HW2, HW2 Solution, Thu, Photo

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Sep 28

Tue, HW3, HW3 Solution, Thu

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Oct 5

Tue, HW4, HW4 Solution, Thu,

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Oct 12

Tue, Thu

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Oct 19

Tue, HW5, HW5 Solution, Term Test on Thu

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Oct 26

Tue, Why LinAlg?, HW6, HW6 Solution, Thu

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Nov 2

Tue, MIT LinAlg, Thu

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Nov 9

Tue, HW7, HW7 Solution Thu

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Nov 16

Tue, HW8, HW8 Solution, Thu

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Nov 23

Tue, HW9, HW9 Solution, Thu

13

Nov 30

Tue, On the final, Thu

S

Dec 7

Office Hours

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Dec 14

Final on Dec 16

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Replacement Theorem
...

dim(V) = n
 If G generates V then $G\geq n$. If also $\,\!G=n$ then G is a basis.
 If L is linearly dependent then $L\leq n$. If also $\,\!L=n$ then L is a basis. If also $\,\!L<n$ 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 $G\geq n$.
a2. Let $\beta$ be a basis of V, then $\,\!B=n$. Now use replacement with G & L = $\beta$. Hence, $G\geq L=B=n$.
a. From a1 and a2, we know $G\geq n$. If $\,\!G=n$ then G contains a basis $\beta$. But $\,\!B=n$, so $\,\!G=\beta$, and hence G is a basis.
b. Use replacement with G' being some basis of V. G = n.
If $L=G$ then $R=n=G$, so $R=G$, so $(G\backslash R)\cup L$ generates, so L generates, so L is a basis since it is linearly independent.
We have that L is basis. If $L<G$ the nagain find $R\subset G$ such that $R=L$ and $(G\backslash R)\cup L$ generates.
1. $\beta$ generates V.
2. $B\leq GR+L=n$. So by part a, $\beta$ is a basis.
3.

If V is finitedimensional (f.d.) and $\mathbf {W} \subset \mathbf {V}$ is a subspace of V, then W is also finite, and $\operatorname {dim} (\mathbf {W} )\leq \operatorname {dim} (\mathbf {V} )$. 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 finitedimensional, pick a basis $\beta$ of W; $\beta$ is linearly independent in V so by Corollary 3 of part b, $B\leq \operatorname {dim} (\mathbf {V} )\Rightarrow \operatorname {dim} (\mathbf {W} )=B\leq \operatorname {dim} (\mathbf {V} )$. So span($\beta$) = V = W so V = W. ...
Assume W is not finitedimensional. $\mathbf {W} \neq \{0\}$ so pick a $x_{1}\in W$ such that $x_{1}\neq 0$. So $\{x_{1}\}$ is linearly independent in W, and $\operatorname {span} (\{x_{1}\})\subsetneq \mathbf {W}$. Pick $x_{2}\in W\backslash \operatorname {span} (\{x_{1}\})$. So $\{x_{1},x_{2}\}$ is linearly dependent and $\operatorname {span} (\{x_{1},x_{2}\})\subsetneq \mathbf {W}$. Pick $x_{3}\in W\backslash \operatorname {span} (\{x_{1},x_{2}\})$ ... continue in this way to get a sequence $x_{1},x_{2},\ldots ,x_{n+1}$ where n = dim(V) and $\{x_{1},\ldots ,x_{n+1}\}$ is linearly independent. There is a contradiction by Corollary 3.b.
The Lagrange Interpolation Formula
[Aside: $(ax)(bx)\ldots (zx)=0$ because $xx=0$]
Where $1\leq i<n,$
Let $x_{i}$ be distinct points in $\Re$.
Let $y_{i}$ be any points in $\Re$.
Can you find a polynomial $P\in \mathbb {P} _{n}(\Re )$ such that $P(x_{i})=y_{i}$? Is it unique?
Example:
 $x_{i}=0,1,3$
 $y_{i}=5,2,2$
Can we find a $P\in \mathbb {P} _{2}$ such that
 $P(0)=5$
 $P(1)=2$
 $P(2)=2$?
Solution: Let ${\tilde {P}}_{i}(x)=\prod _{j=1,j\neq 1}^{n+1}(xx_{j})\in \mathbb {P} _{n}(\Re )$. (Remember capital pi notation.)
Then ${\tilde {P}}_{i}(x_{k})={\begin{cases}0&i\neq k\\\prod _{i\neq j}(x_{i}x_{j})&i=k\\\end{cases}}$
 ${\tilde {P}}_{1}=(xx_{2})(xx_{3})=(x1)(x3)=x^{2}4x+3$
 ${\tilde {P}}_{2}=(x0)(x3)=x^{2}3x$
 ${\tilde {P}}_{3}=(x0)(x1)=x^{2}x$
${\begin{matrix}{\tilde {P}}_{1}(0)=3&{\tilde {P}}_{1}(1)=0&{\tilde {P}}_{1}(3)=0\\{\tilde {P}}_{2}(0)=0&{\tilde {P}}_{2}(1)=2&{\tilde {P}}_{2}(3)=0\\{\tilde {P}}_{3}(0)=0&{\tilde {P}}_{3}(1)=0&{\tilde {P}}_{3}(3)=6\\\end{matrix}}$
Set $P_{i}(x)={\frac {1}{{\tilde {P}}_{i}(x)}}\cdot {\tilde {P}}_{i}=\prod _{j\neq i}{\frac {(xx_{j})}{(x_{i}x_{j})}}$
Then $P_{i}(x_{k})={\begin{cases}0&i\neq k\\1&i=k\\\end{cases}}$
 $P_{1}(x)={\frac {1}{3}}x^{2}{\frac {4}{3}}x+1$
 $P_{2}(x)={\frac {1}{2}}x^{2}+{\frac {3}{2}}x$
 $P_{3}(x)={\frac {1}{6}}x^{2}{\frac {1}{6}}x$
Let $P=\sum _{i=1}^{n+1}y_{i}P_{i}(x)\in \mathbb {P} _{n}(\Re )$
 $P=5\cdot \left({\frac {1}{3}}x^{2}{\frac {4}{3}}x+1\right)+2\cdot \left({\frac {1}{2}}x^{2}+{\frac {3}{2}}x\right)+2\cdot \left({\frac {1}{6}}x^{2}{\frac {1}{6}}x\right)$
 $\,\!=x^{2}4x+5$