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 January 2nd, 2018, 12:43 PM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 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: 311 Thanks: 109 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: 124 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: 753 Thanks: 261 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: 124 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: 753 Thanks: 261 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: 124 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"-."

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