# Is a dichotomic basis possible for 3-dimensional space?

We know that the Pauli basis for the 2-dimensional space is a dichotomic basis in the sense that every Pauli matrix has two distinct eigenvalues. Is it possible to express a 3-dimensional matrix $$\textbf{M}$$ in terms of a basis $$\{\textbf{B}_k\}$$, that is $$\textbf{M} = \sum\limits_k c_k \textbf{B}_k$$ ($$c_k$$ are the coefficients), such that every $$\textbf{B}_k$$ has only two distinct eigenvalues?

## 1 Answer

Every matrix, regardless of dimensionality, is the sum of the projections onto its eigenspaces weighted by their eigenvalues. A projector only has two eigenvalues, 1 and 0. So choose $$c_k$$ as the eigenvalues and $$B_k$$ as the projectors to achieve what you want.

• @ChiralAnomaly Ah, yes, either diagonalizability or non-zero eigenvalues are probably necessary – ACuriousMind Mar 28 at 6:49