# How many eigenbases can a Hermitian matrix have?

I learned in quantum mechanics that the eigenvectors of a Hermitian operator (matrix) can be made to form an orthogonal eigenbasis for the vector space.

I also learned that 2 commuting Hermitian operators share a common eigenbasis. Does this mean that Hermitian operators can have more than one eigenbasis? In general, how many eigenbases can a Hermitian operator have?

## 4 Answers

arbitrarily many, just take the matrix corresponding to the scalar multiplicattion, this has every basis as an eigenbasis

• do you mean if ${ \vec{x}, \vec{y}, \vec{z} }$ form an eigenbasis for a Hermitian operator, ${ A \vec{x}, A \vec{y}, A \vec{z} }$ is another one, where A is a scalar? – TaeNyFan Feb 15 at 15:35
• is it possible to have $\vec u, \vec v, \vec i$ as another eigenbasis, where these are not necessarily ${ A \vec{x}, A \vec{y}, A \vec{z} }$? – TaeNyFan Feb 15 at 15:37

It is not true that two commuting Hermitian operators share a common eigenbasis, as you can see in this question, but if two operators share an eigenbasis, that is, they are simultaneously diagonalizable, they must commute.

If the eigenvalues of a diagonalizable operator $$A$$ all have multiplicity $$1$$ then the eigenbasis is unique up to multiplication of the eigenvalues by scalars. There is essentially just one basis of eigenvectors.

Any operator commuting with $$A$$ will have the same eigenvectors as $$A$$. To see why, suppose $$v$$ is an eigenvector for $$A$$. Then $$ABv = BAv = B\lambda v = \lambda Bv$$ so $$Bv$$ is a $$\lambda$$-eigenvector of $$A$$. Since the space of those eigenvectors is one dimensional and spanned by $$v$$, $$Bv$$ must be a multiple of $$v$$, hence an eigenvector of $$B$$ for some eigenvalue.

I suspect that the multiplicity $$1$$ assumption is implicit in the quantum mechanical application you have in mind.

The identity $$I$$ has a lot eigen bases. Any time an eigenspace has dimension $$> 1$$, there are different eigen bases. If the multiplicities of the eigen bases are all $$1$$, then your basis elements are unique up to unimodular constant multipliers.