# Generalized eigen property of a matrix

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 $$|d_{11}| = \dots = |d_{nn}| = 1$$

• 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 ${\rm det}\,(D-A)=0$. Mar 21, 2019 at 10:08
• @CarloBeenakker Right, I included the eigenvectors to the question.
– Jiro
Mar 21, 2019 at 10:21
• 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). Mar 24, 2019 at 22:35
• @NikWeaver I see this, but how can you generalize from this fact to non-triangular matrices?
– Jiro
Mar 25, 2019 at 8:14
• 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 $S_A = \{(d_{11}, d_{22})|(d_{11} - 0.8)(d_{22} - 0.8) + 0.36 = 0\}$. Mar 29, 2019 at 18:10