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Given a $n \times n$ invertible matrix $A$, I am interested in the set

$$ \mathcal{S}(A) = \{ D \textrm{ diagonal matrix } \mid \det(D - A) = 0 \}. $$

Thus, for all eigenvalues $\lambda_i$, we have $\lambda_i I \in \mathcal{S}(A)$, where $I$ is the identity matrix.

I would like to learn more about this set $\mathcal{S}(A)$. Is there a name for it in literature? In particular, I am interested whether it can be parametrized in terms of the eigenvalues and eigenvectors of $A$. Further, I am interested in unitary $U$ and $ \mathcal{S}_\textrm{1} \subset \mathcal{S}(U)$ such that for $D \in \mathcal{S}_\textrm{1}$ that $$|d_{11}| = \dots = |d_{nn}| = 1$$

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    $\begingroup$ inspection of $n=2$, for a symmetric $A$, will tell you that you need the eigenvectors as well as the eigenvalues of $A$ to parameterize the two diagonal elements of $D$ that satisfy ${\rm det}\,(D-A)=0$. $\endgroup$
    – Carlo Beenakker
    Commented Mar 21, 2019 at 10:08
  • $\begingroup$ @CarloBeenakker Right, I included the eigenvectors to the question. $\endgroup$
    – Jiro
    Commented Mar 21, 2019 at 10:21
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    $\begingroup$ If $A$ is upper triangular then this set consists of all diagonal matrices which share at least one diagonal entry with $A$ (same value in the same position). $\endgroup$
    – Nik Weaver
    Commented Mar 24, 2019 at 22:35
  • $\begingroup$ @NikWeaver I see this, but how can you generalize from this fact to non-triangular matrices? $\endgroup$
    – Jiro
    Commented Mar 25, 2019 at 8:14
  • $\begingroup$ Requiring $A$ to be unitary doesn't restrict $D$ to be unitary. Take, for example, $A = \left[\begin{array}{cc} 0.8 & 0.6 \\ -0.6 & 0.8 \end{array} \right]$; then $S_A = \{(d_{11}, d_{22})|(d_{11} - 0.8)(d_{22} - 0.8) + 0.36 = 0\}$. $\endgroup$
    – user44191
    Commented Mar 29, 2019 at 18:10

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