09-240/Classnotes for Tuesday October 13

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Replacement Theorem

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3. dim(V) = n
   1. If G generates V then $|G|\geq n$. If also $\,\!|G|=n$ then G is a basis.
   2. 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. $|B|\leq |G|-|R|+|L|=n$. So by part a, $\beta$ is a basis.
      3.
4. If V is finite-dimensional (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 finite-dimensional, 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 finite-dimensional. $\mathbf {W} \neq \{0\}$ 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: $(a-x)(b-x)\ldots (z-x)=0$ because $x-x=0$]

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 ${\tilde {P}}_{i}(x)=\prod _{j=1,j\neq 1}^{n+1}(x-x_{j})\in \mathbb {P} _{n}(\Re )$. (Remember capital pi notation.)

Then ${\tilde {P}}_{i}(x_{k})={\begin{cases}0&i\neq k\\\prod _{i\neq j}(x_{i}-x_{j})&i=k\\\end{cases}}$

${\tilde {P}}_{1}=(x-x_{2})(x-x_{3})=(x-1)(x-3)=x^{2}-4x+3$

${\tilde {P}}_{2}=(x-0)(x-3)=x^{2}-3x$

${\tilde {P}}_{3}=(x-0)(x-1)=x^{2}-x$

${\begin{matrix}{\tilde {P}}_{1}(0)=3&{\tilde {P}}_{1}(1)=0&{\tilde {P}}_{1}(3)=0\\{\tilde {P}}_{2}(0)=0&{\tilde {P}}_{2}(1)=-2&{\tilde {P}}_{2}(3)=0\\{\tilde {P}}_{3}(0)=0&{\tilde {P}}_{3}(1)=0&{\tilde {P}}_{3}(3)=6\\\end{matrix}}$

Set $P_{i}(x)={\frac {1}{{\tilde {P}}_{i}(x)}}\cdot {\tilde {P}}_{i}=\prod _{j\neq i}{\frac {(x-x_{j})}{(x_{i}-x_{j})}}$

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 $P=\sum _{i=1}^{n+1}y_{i}P_{i}(x)\in \mathbb {P} _{n}(\Re )$

$P=5\cdot \left({\frac {1}{3}}x^{2}-{\frac {4}{3}}x+1\right)+2\cdot \left(-{\frac {1}{2}}x^{2}+{\frac {3}{2}}x\right)+2\cdot \left({\frac {1}{6}}x^{2}-{\frac {1}{6}}x\right)$

$\,\!=x^{2}-4x+5$