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 January 2nd, 2018, 12:43 PM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Eigenvectors of a distinct eigenvalue Why is there only one eigenvector (up to scalar multiplication) corresponding to a distinct eigenvalue? January 2nd, 2018, 01:14 PM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry That's not true. For example, take the identity matrix - every vector is an eigenvector for the eigenvalue 1. January 2nd, 2018, 03:29 PM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I said distinct eigenvalue, not repeated eigenvalues which can have multiple eigenvectors. January 2nd, 2018, 03:48 PM #4 Senior Member   Joined: Oct 2009 Posts: 866 Thanks: 329 I don't understand the question then. What is a distinct eigenvalue? Don't you need two things in order for something to be distinct? January 2nd, 2018, 04:08 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Examples of eigenvalues of a 3x3 matrix: distinct eigenvalues: 2,5,7 repeated eigenvalues: 3,3,5 January 2nd, 2018, 04:59 PM   #6
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Quote:
 Originally Posted by zylo Why is there only one eigenvector (up to scalar multiplication) corresponding to a distinct eigenvalue?
If you mean that we get exactly $n$ distict eigenvalues $\lambda_1,\cdots,\lambda_n$ of a $n\times n$ square matrix where $\lambda_i\ne\lambda_j$ for $i\ne j$, in this case the dimension of the sub-vector space of the eigenvectors of each eigenvalue is 1. January 2nd, 2018, 08:42 PM #7 Senior Member   Joined: Oct 2009 Posts: 866 Thanks: 329 Or said differently: it is because eigenvectors corresponding to different eigenvalues are linearly independent. So the existence of n distinct eigenvalues implies that there are n linearly independent eigenvectors each coming from a different eigenvalue. So there is just not enough place for a second independent eigenvector for a given eigenvalue. January 5th, 2018, 01:22 PM #8 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Found a Theorem.* For any nxn matrix, R + N = n R = rank, (number of linearly independent columns) N = dimension of null space, If R = n, N = 0 If R = n-1, N=1 Eigenvectors x of eigenvalue $\displaystyle \lambda$ are given by $\displaystyle [A - \lambda I]x = 0$ and N is dimension of null space (eigenspace). Algebraic multiplicity m$\displaystyle _{\lambda}$: number of repetitions of $\displaystyle \lambda$, Geometric multiplicity: number of eigenvectors of $\displaystyle \lambda$ Rank Multiplicity Theorem* $\displaystyle R[A - \lambda I] \geq n=m_{\lambda}$ If $\displaystyle m_{\lambda} = 1$, ($\displaystyle \lambda$ is distinct), R = n-1 and N = 1, i.e., there is only one eigenvector of a distinct eigenvalue. Since $\displaystyle R \geq n - m_{\lambda}$ , $\displaystyle N=n-R \leq m_{\lambda}$, or Geometric multiplicity $\displaystyle \leq$ Algebraic multiplicity. *Mirsky, Intro to LA, pg 214 Last edited by skipjack; January 5th, 2018 at 02:03 PM. January 6th, 2018, 08:12 AM   #9
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Quote:
 Originally Posted by zylo Found a Theorem.* Rankk Multiplicity Theorem* $\displaystyle R[A - \lambda I] \geq n=m_{\lambda}$ *Mirsky, Intro to LA, pg 214
The "=" should be "-."

$\displaystyle R[A - \lambda I] \geq n-m_{\lambda}$

Seems to me if your going to edit(?} a post it would be nice to correct an obvious typo.

Last edited by zylo; January 6th, 2018 at 08:18 AM. Reason: Add space in be"-." Tags distinct, eigenvalue, eigenvectors Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post whitegreen Linear Algebra 1 August 17th, 2016 06:55 AM dnopas Linear Algebra 1 October 22nd, 2015 12:59 PM TomCadwallader Linear Algebra 2 November 17th, 2013 05:20 PM metamath101 Algebra 2 June 22nd, 2012 04:59 PM tinng Applied Math 1 December 15th, 2009 12:08 PM

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