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January 2nd, 2018, 01:43 PM   #1
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Eigenvectors of a distinct eigenvalue

Why is there only one eigenvector (up to scalar multiplication) corresponding to a distinct eigenvalue?
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January 2nd, 2018, 02:14 PM   #2
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That's not true. For example, take the identity matrix - every vector is an eigenvector for the eigenvalue 1.
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January 2nd, 2018, 04:29 PM   #3
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I said distinct eigenvalue, not repeated eigenvalues which can have multiple eigenvectors.
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January 2nd, 2018, 04:48 PM   #4
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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?
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January 2nd, 2018, 05:08 PM   #5
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Examples of eigenvalues of a 3x3 matrix:
distinct eigenvalues: 2,5,7
repeated eigenvalues: 3,3,5
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January 2nd, 2018, 05:59 PM   #6
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Salam !

Quote:
Originally Posted by zylo View Post
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.
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January 2nd, 2018, 09:42 PM   #7
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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.
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January 5th, 2018, 02:22 PM   #8
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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 03:03 PM.
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January 6th, 2018, 09:12 AM   #9
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Quote:
Originally Posted by zylo View Post
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 09:18 AM. Reason: Add space in be"-."
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