09-240/Classnotes for Tuesday October 13

09-240/Navigation Panel   Additions to the MAT 240 web site no longer count towards good deed points # Week of... Notes and Links 1 Sep 7 Tue, About, Thu 2 Sep 14 Tue, HW1, HW1 Solution, Thu 3 Sep 21 Tue, HW2, HW2 Solution, Thu, Photo 4 Sep 28 Tue, HW3, HW3 Solution, Thu 5 Oct 5 Tue, HW4, HW4 Solution, Thu, 6 Oct 12 Tue, Thu 7 Oct 19 Tue, HW5, HW5 Solution, Term Test on Thu 8 Oct 26 Tue, Why LinAlg?, HW6, HW6 Solution, Thu 9 Nov 2 Tue, MIT LinAlg, Thu 10 Nov 9 Tue, HW7, HW7 Solution Thu 11 Nov 16 Tue, HW8, HW8 Solution, Thu 12 Nov 23 Tue, HW9, HW9 Solution, Thu 13 Nov 30 Tue, On the final, Thu S Dec 7 Office Hours F Dec 14 Final on Dec 16 To Do List The Algebra Song! Register of Good Deeds Misplaced Material Add your name / see who's in!

WARNING: The notes below, written for students and by students, are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them. It is a bad idea to stop taking your own notes thinking that these notes can be a total replacement - there's nothing like one's own handwriting! Visit this pages' history tab to see who added what and when.

• 

  Bright - 1

• 

  2

• 

  3

• 

  4

Replacement Theorem

...

3. dim(V) = n
   1. If G generates V then [math]\displaystyle{ |G| \ge n }[/math]. If also [math]\displaystyle{ \,\! |G| = n }[/math] then G is a basis.
   2. If L is linearly dependent then [math]\displaystyle{ |L| \le n }[/math]. If also [math]\displaystyle{ \,\! |L| = n }[/math] then L is a basis. If also [math]\displaystyle{ \,\! |L| \lt n }[/math] 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 [math]\displaystyle{ |G| \ge n }[/math].
      a2. Let [math]\displaystyle{ \beta }[/math] be a basis of V, then [math]\displaystyle{ \,\! |B| = n }[/math]. Now use replacement with G & L = [math]\displaystyle{ \beta }[/math]. Hence, [math]\displaystyle{ |G| \ge |L| = |B| = n }[/math].
      a. From a1 and a2, we know [math]\displaystyle{ |G| \ge n }[/math]. If [math]\displaystyle{ \,\! |G| = n }[/math] then G contains a basis [math]\displaystyle{ \beta }[/math]. But [math]\displaystyle{ \,\! |B| = n }[/math], so [math]\displaystyle{ \,\! |G| = \beta }[/math], and hence G is a basis.
      
      b. Use replacement with G' being some basis of V. |G| = n.
      
      If [math]\displaystyle{ |L| = |G| }[/math] then [math]\displaystyle{ |R| = n = |G| }[/math], so [math]\displaystyle{ R = G }[/math], so [math]\displaystyle{ (G \backslash R) \cup L }[/math] generates, so L generates, so L is a basis since it is linearly independent.
      
      We have that L is basis. If [math]\displaystyle{ |L| \lt |G| }[/math] the nagain find [math]\displaystyle{ R \subset G }[/math] such that [math]\displaystyle{ |R| = |L| }[/math] and [math]\displaystyle{ (G \backslash R) \cup L }[/math] generates.
      
      1. [math]\displaystyle{ \beta }[/math] generates V.
      2. [math]\displaystyle{ |B| \le |G| - |R| + |L| = n }[/math]. So by part a, [math]\displaystyle{ \beta }[/math] is a basis.
      3.
4. If V is finite-dimensional (f.d.) and [math]\displaystyle{ \mathbf W \subset \mathbf V }[/math] is a subspace of V, then W is also finite, and [math]\displaystyle{ \operatorname{dim}(\mathbf W) \le \operatorname{dim}(\mathbf V) }[/math]. 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 [math]\displaystyle{ \beta }[/math] of W; [math]\displaystyle{ \beta }[/math] is linearly independent in V so by Corollary 3 of part b, [math]\displaystyle{ |B| \le \operatorname{dim}(\mathbf V) \Rightarrow \operatorname{dim}(\mathbf W) = |B| \le \operatorname{dim}(\mathbf V) }[/math]. So span([math]\displaystyle{ \beta }[/math]) = V = W so V = W. ...
   
   Assume W is not finite-dimensional. [math]\displaystyle{ \mathbf W \ne \{0\} }[/math] so pick a [math]\displaystyle{ x_1 \in W }[/math] such that [math]\displaystyle{ x_1 \ne 0 }[/math]. So [math]\displaystyle{ \{x_1\} }[/math] is linearly independent in W, and [math]\displaystyle{ \operatorname{span}(\{x_1\}) \subsetneq \mathbf W }[/math]. Pick [math]\displaystyle{ x_2 \in W \backslash \operatorname{span}(\{x_1\}) }[/math]. So [math]\displaystyle{ \{x_1, x_2\} }[/math] is linearly dependent and [math]\displaystyle{ \operatorname{span}(\{x_1, x_2\}) \subsetneq \mathbf W }[/math]. Pick [math]\displaystyle{ x_3 \in W \backslash \operatorname{span}(\{x_1, x_2\}) }[/math] ... continue in this way to get a sequence [math]\displaystyle{ x_1, x_2, \ldots, x_{n+1} }[/math] where n = dim(V) and [math]\displaystyle{ \{x_1, \ldots, x_{n+1}\} }[/math] is linearly independent. There is a contradiction by Corollary 3.b.

The Lagrange Interpolation Formula

[Aside: [math]\displaystyle{ (a - x)(b - x) \ldots (z - x) = 0 }[/math] because [math]\displaystyle{ x - x = 0 }[/math]]

Where [math]\displaystyle{ 1 \le i \lt n, }[/math]

Let [math]\displaystyle{ x_i }[/math] be distinct points in [math]\displaystyle{ \real }[/math].

Let [math]\displaystyle{ y_i }[/math] be any points in [math]\displaystyle{ \real }[/math].

Can you find a polynomial [math]\displaystyle{ P \in \mathbb P_n(\real) }[/math] such that [math]\displaystyle{ P(x_i) = y_i }[/math]? Is it unique?

Example:

[math]\displaystyle{ x_i = 0, 1, 3 }[/math]

[math]\displaystyle{ y_i = 5, 2, 2 }[/math]

Can we find a [math]\displaystyle{ P \in \mathbb P_2 }[/math] such that

[math]\displaystyle{ P(0) = 5 }[/math]

[math]\displaystyle{ P(1) = 2 }[/math]

[math]\displaystyle{ P(2) = 2 }[/math]?

Solution: Let [math]\displaystyle{ \tilde P_i(x) = \prod_{j=1, j \ne 1}^{n+1} (x - x_j) \in \mathbb P_n(\real) }[/math]. (Remember capital pi notation.)

Then [math]\displaystyle{ \tilde P_i(x_k) = \begin{cases} 0 & i \ne k \\ \prod_{i \ne j} (x_i - x_j) & i = k \\ \end{cases} }[/math]

[math]\displaystyle{ \tilde P_1 = (x - x_2)(x - x_3) = (x - 1)(x - 3) = x^2 - 4x + 3 }[/math]

[math]\displaystyle{ \tilde P_2 = (x - 0)(x - 3) = x^2 - 3x }[/math]

[math]\displaystyle{ \tilde P_3 = (x - 0)(x - 1) = x^2 - x }[/math]

[math]\displaystyle{ \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} }[/math]

Set [math]\displaystyle{ P_i(x) = \frac1{\tilde P_i(x)} \cdot \tilde P_i = \prod_{j \ne i} \frac{(x - x_j)}{(x_i - x_j)} }[/math]

Then [math]\displaystyle{ P_i(x_k) = \begin{cases} 0 & i \ne k \\ 1 & i = k \\ \end{cases} }[/math]

[math]\displaystyle{ P_1(x) = \frac13 x^2 - \frac43 x + 1 }[/math]

[math]\displaystyle{ P_2(x) = -\frac12 x^2 + \frac32 x }[/math]

[math]\displaystyle{ P_3(x) = \frac16 x^2 - \frac16 x }[/math]

Let [math]\displaystyle{ P = \sum_{i=1}^{n+1} y_i P_i(x) \in \mathbb P_n(\real) }[/math]

[math]\displaystyle{ P = 5 \cdot \left( \frac13 x^2 - \frac43 x + 1 \right) + 2 \cdot \left( -\frac12 x^2 + \frac32 x \right) + 2 \cdot \left( \frac16 x^2 - \frac16 x \right) }[/math]

[math]\displaystyle{ \,\! = x^2 - 4x + 5 }[/math]