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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?

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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.

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

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