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Repeatedly applying the previous product rule givesFormally, the result holds for $p=1$; multiplication by $B$ preserves the zero off-diagonal matrix entries and multiplies each diagonal matrix entry by $B_{ii}$, proving the induction step.

Let $e_{ij}$ denote the matrix unit with one at position $(i,j)$ and zero elsewhere. The five inverses areThus each is the identity matrix with only the indicated off-diagonal matrix entry altered. These inverses reverse the corresponding row additions; no row scaling or permutation is involved.

Three equally weighted pure qubit states with coplanar unit Bloch vectors separated by $120$ degrees. One example is $|t_j⟩=(|0⟩+{\omega }^j|1⟩)/\sqrt{2}$, with $\omega =e^{2\pi i/3}$ and $j=0,1,2$. The roots-of-unity identity $1+\omega +{\omega }^2=0$ cancels the average off-diagonal matrix entries, so $\frac{1}{3}{\sum }_j|t_j⟩⟨t_j|=I/2$.

This is an ensemble of preparation states, whereas a trine qubit POVM consists of measurement effects $E_j=\frac{2}{3}|t_j⟩⟨t_j|$. The POVM–ensemble duality for the maximally mixed state relates them but does not identify their operational roles.