I'm interested in the case of a specific matrix having different eigenvectors corresponding to two identical eigenvalues. The method I use for spectral decomposition returns different eigenvectors, even though the eigenvalue is the same. Is this possible, and if so, what this tells about the matrix?
Same eigenvalues, different eigenvectors
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1Any vector is an eigenvector with eigenvalue $1$ for the identity matrix. All eigenvectors with a given eigenvalue form a linear space, so there will never be just one. – 2012-03-16
4 Answers
Of course it's possible: $ \begin{bmatrix} 2&0\\ 0&2 \end{bmatrix} \, \begin{bmatrix} 1\\ 0 \end{bmatrix} = 2\;\begin{bmatrix} 1\\ 0 \end{bmatrix}, \ \ \ \ \ \ \begin{bmatrix} 2&0\\ 0&2 \end{bmatrix} \, \begin{bmatrix} 0\\ 1 \end{bmatrix} = 2\;\begin{bmatrix} 0\\ 1 \end{bmatrix}. $
What it tells you about the matrix is that the geometric multiplicity of the eigenvalue is greater than $1$.
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0What do you mean by geometric multiplicity? How is different from algebraic multiplicity? – 2016-07-15
A trivial example: Consider the 2 by 2 identity matrix. It has only one eigenvalue, namely 1. However both $e_1=(1,0)$ and $e_2=(0,1)$ are eigenvectors of this matrix.
Quick Test of 2x2 matrix where a are eigenvalues (same).
| a b |
| 0 a |
if b = 0 there are 2 different eigenvectors for same eigenvalue a
if b != 0 (not equal) then there is only one eigenvector for eigenvalue a
Every eigenvalue with multiplicity = n will be associated with n different (as in linearly independent) eigenvalues.
Multiplicity is how many "times" it shows up as an eigenvalue. It is like when you find only one solution to a second degree equation, which always has two roots. This solution has a multiplicity = 2.