# Same eigenvalues, different eigenvectors

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?

• 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. Mar 16, 2012 at 19:09

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 $1$.

• What do you mean by geometric multiplicity? How is different from algebraic multiplicity? Jul 15, 2016 at 7:17
• (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. Sep 17, 2020 at 17:30
• 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.
– MSIS
Sep 25, 2023 at 23:36
• 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 $S$. Sep 26, 2023 at 1:06

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.

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.

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 $$a$$.