09-240/Classnotes for Tuesday October 13

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Replacement Theorem

...

  1. dim(V) = n
    1. If G generates V then |G| \ge n. If also \,\! |G| = n then G is a basis.
    2. If L is linearly dependent then |L| \le 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| \ge n.
      a2. Let \beta be a basis of V, then \,\! |B| = n. Now use replacement with G & L = \beta. Hence, |G| \ge |L| = |B| = n.
      a. From a1 and a2, we know |G| \ge 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| \le |G| - |R| + |L| = n. So by part a, \beta is a basis.
      3.
  2. 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) \le \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| \le \operatorname{dim}(\mathbf V) \Rightarrow \operatorname{dim}(\mathbf W) = |B| \le \operatorname{dim}(\mathbf V). So span(\beta) = V = W so V = W. ...

    Assume W is not finite-dimensional. \mathbf W \ne \{0\} so pick a x_1 \in W such that x_1 \ne 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 \le i < n,

Let x_i be distinct points in \real.

Let y_i be any points in \real.

Can you find a polynomial P \in \mathbb P_n(\real) 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 \ne 1}^{n+1} (x - x_j) \in \mathbb P_n(\real). (Remember capital pi notation.)

Then \tilde P_i(x_k) = \begin{cases}
0 & i \ne k \\
\prod_{i \ne 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) = \frac1{\tilde P_i(x)} \cdot \tilde P_i = \prod_{j \ne i} \frac{(x - x_j)}{(x_i - x_j)}

Then P_i(x_k) = \begin{cases}
0 & i \ne k \\
1 & i = k \\
\end{cases}

P_1(x) = \frac13 x^2 - \frac43 x + 1
P_2(x) = -\frac12 x^2 + \frac32 x
P_3(x) = \frac16 x^2 - \frac16 x

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

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)
\,\! = x^2 - 4x + 5