06-240/Classnotes For Tuesday December 5: Difference between revisions

Revision as of 12:16, 5 December 2006

#	Week of...	Notes and Links
1	Sep 11	About, Tue, HW1, Putnam, Thu
2	Sep 18	Tue, HW2, Thu
3	Sep 25	Tue, HW3, Photo, Thu
4	Oct 2	Tue, HW4, Thu
5	Oct 9	Tue, HW5, Thu
6	Oct 16	Why?, Iso, Tue, Thu
7	Oct 23	Term Test, Thu (double)
8	Oct 30	Tue, HW6, Thu
9	Nov 6	Tue, HW7, Thu
10	Nov 13	Tue, HW8, Thu
11	Nov 20	Tue, HW9, Thu
12	Nov 27	Tue, HW10, Thu
13	Dec 4	On the final, Tue, Thu
F	Dec 11	Final: Dec 13 2-5PM at BN3, Exam Forum
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Our remaining goal for this semester is to study the following theorem:

Theorem. Let $A$ be an $n\times n$ matrix (with entries in some field $F$ ) and let $\chi _{A}(\lambda ):=\det(A-\lambda I)$ be the characteristic polynomial of $A$ . Assume $\chi _{A}$ has $n$ distinct roots $\lambda _{1}\ldots \lambda _{n}$ , that is, $A$ has $n$ distinct eigenvalues $\lambda _{1}\ldots \lambda _{n}$ , and let $v_{1},\ldots ,v_{n}$ be corresponding eigenvectors, so that $Av_{i}=\lambda _{i}v_{i}$ for all $1\leq i\leq n$ . Let $D$ be the diagonal matrix that has $\lambda _{1}$ through $\lambda _{n}$ on its main diagonal (in order) and let $P$ be the matrix whose columns are these eigenvectors: $P:=(v_{1}|v_{2}|\cdots |v_{n})$ . Then $P$ is invertible and the following equalities hold:

$D=P^{-1}AP$ and $A=PDP^{-1}$ .
For any positive integer $k$ we have $A^{k}=PD^{k}P^{-1}$ and $D^{k}={\begin{pmatrix}\lambda _{1}^{k}&&0\\&\ddots &\\0&&\lambda _{n}^{k}\end{pmatrix}}$ .
Likewise if $F={\mathbb {R} }$ and $\exp(B):=\sum _{k=0}^{\infty }{\frac {B^{k}}{k!}}$ then $\exp(A)=P\exp(D)P^{-1}$ and $\exp(D)={\begin{pmatrix}e^{\lambda _{1}}&&0\\&\ddots &\\0&&e^{\lambda _{n}}\end{pmatrix}}$ .

Order of the proceedings.

Assuming P is invertible, a proof of 1.
Proof of 2.
Example - the "reproduction of rabbits" matrix $A={\begin{pmatrix}0&1\\1&1\end{pmatrix}}$ (see the mathematica session below).
Discussion of 3.
The relationship with linear transformations and changes of basis.
$v_{1}$ thorough $v_{n}$ form a basis and $P$ is invertible.

@@ Line 1: / Line 1: @@
 {{06-240/Navigation}}
-{{In Preparation}}
 Our remaining goal for this semester is to study the following theorem:
-'''Theorem.''' Let <math>A</math> be an <math>n\times n</math> matrix (with entries in some field <math>F</math>) and let <math>\chi(\lambda):=\det(A-\lambda I)</math> be the characteristic polynomial of <math>A</math>. Assume <math>\chi</math> has <math>n</math> distinct roots <math>\lambda_1\ldots\lambda_n</math>, that is, <math>A</math> has <math>n</math> distinct eigenvalues <math>\lambda_1\ldots\lambda_n</math>, and let <math>v_1,\ldots,v_n</math> be corresponding eigenvectors, so that <math>Av_i=\lambda_iv_i</math> for all <math>1\leq i\leq n</math>. Let <math>D</math> be the diagonal matrix that has <math>\lambda_1</math> through <math>\lambda_n</math> on its main diagonal (in order) and let <math>P</math> be the matrix whose columns are these eigenvectors: <math>P:=(v_1|v_2|\cdots|v_n)</math>. Then <math>P</math> is invertible and the following equalities hold:
+'''Theorem.''' Let <math>A</math> be an <math>n\times n</math> matrix (with entries in some field <math>F</math>) and let <math>\chi_A(\lambda):=\det(A-\lambda I)</math> be the characteristic polynomial of <math>A</math>. Assume <math>\chi_A</math> has <math>n</math> distinct roots <math>\lambda_1\ldots\lambda_n</math>, that is, <math>A</math> has <math>n</math> distinct eigenvalues <math>\lambda_1\ldots\lambda_n</math>, and let <math>v_1,\ldots,v_n</math> be corresponding eigenvectors, so that <math>Av_i=\lambda_iv_i</math> for all <math>1\leq i\leq n</math>. Let <math>D</math> be the diagonal matrix that has <math>\lambda_1</math> through <math>\lambda_n</math> on its main diagonal (in order) and let <math>P</math> be the matrix whose columns are these eigenvectors: <math>P:=(v_1|v_2|\cdots|v_n)</math>. Then <math>P</math> is invertible and the following equalities hold:
-* <math>D=P^{-1}AP</math> and <math>A=PDP^{-1}</math>.
+# <math>D=P^{-1}AP</math> and <math>A=PDP^{-1}</math>.
-* For any positive integer <math>k</math> we have <math>A^k=PD^kP^{-1}</math> and <math>D^k=\begin{pmatrix}\lambda_1^k&&0\\&\ddots&\\0&&\lambda_n^k\end{pmatrix}</math>.
+# For any positive integer <math>k</math> we have <math>A^k=PD^kP^{-1}</math> and <math>D^k=\begin{pmatrix}\lambda_1^k&&0\\&\ddots&\\0&&\lambda_n^k\end{pmatrix}</math>.
-* Likewise if <math>F={\mathbb R}</math> and <math>\exp(B):=\sum_{k=0}^\infty\frac{B^k}{k!}</math> then <math>\exp(A)=P\exp(D)P^{-1}</math> and <math>\exp(D)=\begin{pmatrix}e^{\lambda_1}&&0\\&\ddots&\\0&&e^{\lambda_n}\end{pmatrix}</math>.
+# Likewise if <math>F={\mathbb R}</math> and <math>\exp(B):=\sum_{k=0}^\infty\frac{B^k}{k!}</math> then <math>\exp(A)=P\exp(D)P^{-1}</math> and <math>\exp(D)=\begin{pmatrix}e^{\lambda_1}&&0\\&\ddots&\\0&&e^{\lambda_n}\end{pmatrix}</math>.
+'''Order of the proceedings.'''
+# Assuming P is invertible, a proof of 1.
+# Proof of 2.
+# Example - the "reproduction of rabbits" matrix <math>A=\begin{pmatrix}0&1\\1&1\end{pmatrix}</math> (see the mathematica session below).
+# Discussion of 3.
+# The relationship with linear transformations and changes of basis.
+# <math>v_1</math> thorough <math>v_n</math> form a basis and <math>P</math> is invertible.
-[[Image:06-240-Reproduction of Rabbits.png|400px]]
+[[Image:06-240-Reproduction of Rabbits.png|center|640px]]

06-240/Classnotes For Tuesday December 5: Difference between revisions

Revision as of 12:16, 5 December 2006

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