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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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    $\begingroup$ 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. $\endgroup$
    – ShawnD
    Mar 16, 2012 at 19:09

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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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  • $\begingroup$ What do you mean by geometric multiplicity? How is different from algebraic multiplicity? $\endgroup$ Jul 15, 2016 at 7:17
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    $\begingroup$ (four years later; I don't know why I didn't reply back then). Geometric multiplicity for an eigenvalue is the dimension of its corresponding eigenspace. Algebraic multiplicity is the order of the eigenvalue as a root of the characteristic polynomial. $\endgroup$ Sep 17, 2020 at 17:30
  • $\begingroup$ I believe the question was whether 2 linear transformations T,T', could have equal eigenvalues, but different Eigenvectors. Even if it wasn't, can you please answer it. $\endgroup$
    – MSIS
    Sep 25, 2023 at 23:36
  • $\begingroup$ Yes, of course. Consider $$\begin{bmatrix} 2&0\\0&0\end{bmatrix}\qquad\text{ and }\qquad \begin{bmatrix} 1&1\\1&1\end{bmatrix}. $$ Both have eigenvalues $0$ and $2$, but the eigenvectors of the first one are (scalar multiplies of) $$\begin{bmatrix} 1\\0\end{bmatrix},\qquad \begin{bmatrix} 0\\1\end{bmatrix},$$ while the second one has eigenvectors (scalar multiples of) $$\begin{bmatrix} 1\\1\end{bmatrix},\qquad \begin{bmatrix} 1\\-1\end{bmatrix}.$$ More generally, $T$ and $STS^{-1}$ have the same eigenvalues for any invertible $S$. $\endgroup$ Sep 26, 2023 at 1:06
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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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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.

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Quick test for a $2 \times 2$ matrix where $a$ are (same) eigenvalues:

\begin{bmatrix} a & b \\ 0 & a\end{bmatrix}.

If $b = 0$, there are $2$ different eigenvectors for same eigenvalue $a$.

If $b \neq 0$, then there is only one eigenvector for eigenvalue $a$.

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