I've seen the state of a quantum system or measurement expressed in a form that implies that all eigenvalues are equal to 1, or can be made so. For example the measurement with $\hat{X}$ of a (discrete) quantum system $\mathbf{S}$ by an observer $\mathbf{A}$ in which the combined system undergoes a process of the form


or a quantum computation in which $ZYX\left|\psi\right>$ expressed as


where, in both cases, $\left|x_i\right>$, $\left|y_j\right>$, $\left|z_k\right>$ are eigenvectors (eigenstates) and the corresponding eigenvalues, $x_i$, $y_j$, $z_k$, that I expect to see as coefficients acquired from the spectral decomposition of the corresponding (unitary, and thus, normal) operator and this associated each term of the form $\left<x_i|\,\cdot\,\right>$, $\left<y_j|\,\cdot\,\right>$, $\left<z_k|\,\cdot\,\right>$ are missing (effectively equalling 1).

Is there some transformation of the form


that satisfies

$$x_i\left|x_i\right>\left<x_i\right|=\left|x_i'\right>\left<x_i'\right|=1\left|x_i'\right>\cdot \left|x_i'\right>^\dagger$$

and that can be applied to any terms were $x_1 \neq 1$ while preserving orthagonality of the $\left|x_i\right>$? If this is possible, what are the conditions (operator properties) that allow it; and if it's not generally possible, what allows one to write the expressions above with (or as if they have) "all-1" eigenvalues, or to simply omit them?

For example the operator $\sigma_1\otimes\sigma_2\otimes\sigma_3$ has several $-1$ eigenvalues. What I can do to the corresponding eigenvectors, such as

$$\begin{bmatrix} i \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix}$$

to "give them" $1$ eigenvalues, or at least give their corresponding outer product terms $1$ weights — or whether this is even possible.

Just to be clear, in the quantum computation case, for example, I expect


I understand that sincesince $X$, $Y$, $Z$ are unitary, $x_i^*x_i = 1$ etc., the eigenvalues don't matter in calculating measurement "probabilities" ($z_k^*y_j^*x_i^*z_ky_jx_i = 1$); but they still should be in the description of the state. No?


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