06-240/Classnotes For Tuesday December 5: Difference between revisions
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'''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: |
'''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: |
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# <math>D=P^{-1}AP</math> and <math>A=PDP^{-1}</math>. |
# <math>D=P^{-1}AP</math> and <math>A=PDP^{-1}</math>. |
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# 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 |
# 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> |
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# 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>. |
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'''Order of the proceedings.''' |
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# Assuming P is invertible, a proof of 1. |
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# Proof of 2. |
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# Example - the "reproduction of rabbits" matrix <math>A=\begin{pmatrix}0&1\\1&1\end{pmatrix}</math> (see the mathematica session below). |
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# Discussion of 3. |
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# The relationship with linear transformations and changes of basis. |
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# <math>v_1</math> thorough <math>v_n</math> form a basis and <math>P</math> is invertible. |
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[[Image:06-240-Reproduction of Rabbits.png|center|640px]] |
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Revision as of 01:37, 28 May 2007
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Our remaining goal for this semester is to study the following theorem:
Theorem. Let be an matrix (with entries in some field ) and let be the characteristic polynomial of . Assume has distinct roots , that is, has distinct eigenvalues , and let be corresponding eigenvectors, so that for all . Let be the diagonal matrix that has through on its main diagonal (in order) and let be the matrix whose columns are these eigenvectors: . Then is invertible and the following equalities hold:
- and .
- For any positive integer we have and Failed to parse (unknown function "\begin{pmatrix}"): {\displaystyle D^k=\begin{pmatrix}\lambda_1^k}
