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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?

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Any 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. – ShawnD Mar 16 '12 at 19:09
up vote 5 down vote accepted

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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What do you mean by geometric multiplicity? How is different from algebraic multiplicity? – Abhishek Bhatia Jul 15 at 7:17

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.

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