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January 2nd, 2018, 12:43 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  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: 211 Thanks: 64 Math Focus: Algebraic Number Theory, Arithmetic 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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
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: 436 Thanks: 147 
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Examples of eigenvalues of a 3x3 matrix: distinct eigenvalues: 2,5,7 repeated eigenvalues: 3,3,5 
January 2nd, 2018, 04:59 PM  #6 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography 
Salam ! 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 subvector space of the eigenvectors of each eigenvalue is 1. 
January 2nd, 2018, 08:42 PM  #7 
Senior Member Joined: Oct 2009 Posts: 436 Thanks: 147 
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
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 = n1, 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 = n1 and N = 1, i.e., there is only one eigenvector of a distinct eigenvalue. Since $\displaystyle R \geq n  m_{\lambda}$ , $\displaystyle N=nR \leq m_{\lambda}$, 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  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  Quote:
$\displaystyle R[A  \lambda I] \geq nm_{\lambda}$ 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"."  

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