06-240/Classnotes For Tuesday December 5 - Revision history

Drorbn: Reverted edit of 213.185.1.179, changed back to last version by Drorbn

2007-05-28T14:50:18Z

Reverted edit of 213.185.1.179, changed back to last version by Drorbn

@@ Line 5: / Line 5: @@
 '''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>.
-# 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</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>.

213.185.1.179 at 06:37, 28 May 2007

2007-05-28T06:37:17Z

@@ Line 5: / Line 5: @@
 '''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>.
-# 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</math>

Drorbn at 16:16, 5 December 2006

2006-12-05T16:16:55Z

← Older revision		Revision as of 12:16, 5 December 2006
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	{{In Preparation}}

	Our remaining goal for this semester is to study the following theorem:		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]]

Drorbn at 16:02, 5 December 2006

2006-12-05T16:02:54Z

← Older revision		Revision as of 12:02, 5 December 2006
Line 1:		Line 1:
	{{06-240/Navigation}}		{{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:
			* <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>.
			* 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>.

	[[Image:06-240-Reproduction of Rabbits.png\|400px]]		[[Image:06-240-Reproduction of Rabbits.png\|400px]]

Drorbn at 03:23, 5 December 2006

2006-12-05T03:23:05Z

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