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Given real vectors $d = (d_1, \ldots, d_n)$ and $\lambda = (\lambda_1, \ldots, \lambda_n)$, where I will assume that their coefficients are arranged in non-increasing order, the Schur-Horn theorem says that there exists a Hermitian matrix with the diagonal given by $d$ and spectrum given by $\lambda$ if and only if $d \prec \lambda$, with $\prec$ denoting majorization (i.e. $ \sum_{i=1}^n d_i = \sum_{i=1}^n \lambda_i$, and $ \sum_{i=1}^k d_i \leq \sum_{i=1}^k \lambda_i$ for all $1 \leq k < n$). Given a vector of eigenvalues, the Schur-Horn theorem then tells us all of the possible diagonal values that Hermitian matrices with such eigenvalues can take. However, it does not take into consideration the eigenvectors of the matrices, and the subspace which they lie in. My question is about an extension of the theorem which does.

Given a vector of eigenvalues $\lambda = (\lambda_1, \ldots, \lambda_r, 0, \ldots, 0)$ and a subspace $S$ representing the span of the eigenvectors corresponding to non-zero eigenvalues, what diagonal values can Hermitian matrices with this set of eigenvalues and with this column space take?

Does this set admit an easy characterization?

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  • $\begingroup$ Does the subspace $S$ correspond to the non-zero eigenvalues? $\endgroup$ – Mahdi May 11 '18 at 12:46
  • $\begingroup$ @Mahdi Yes, sorry I wasn't clear about this! $\endgroup$ – Dario May 11 '18 at 13:13
  • $\begingroup$ Let the columns of $U_1$ be an orthonormal basis for $S$. Then $A=U_1\mathrm{diag}(\lambda_1,\ldots,\lambda_r)U_1^T$ is a typical matrix with the desired property. On the other hand, diagonal entries of $A$ are equal to $u_i^T\mathrm{diag}(\lambda_1,\ldots,\lambda_r)u_i$ where $u_i$s are rows of $U_1$. This give a (not very easy) characterization. $\endgroup$ – Mahdi May 11 '18 at 13:39

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