Let $A,B\in M_n(\mathbb R)$, and $\ell$ a list of $n$ numbers sorted in some order (say, decreasing).
Let $\lambda_i(A)$ be the $i$th eigenvalue of $A$ with respect to the chosen order.
Finally, let $d(A,\ell) = ||\lambda(A)-\ell||_1 = \sum_{i = 1}^n |\lambda_i(A)-\ell_i|$ be the taxicab metric.
My question is as follows:
Given, $A,B,\ell$, is it possible to determine which of $d(A,\ell)$ or $d(B,\ell)$ is smaller without computing the eigenvalues of $A$ and $B$?
Note: The $d(A,\ell)$ can also be any other meaningful metric that determines the matrix which its eigenvalues are closer to the list $\ell$
Update: My solution (Not working!)
Let $\lambda$ and $\beta$ be the eigenvalues of $A$ and $B$ respectively. Clearly $det(A-\lambda I) = 0$ and $det(B-\beta I) = 0$ and $P_n(\lambda), P_n(\beta)$ are characteristic polynomial of $A,B$.
Let sort the eigenvalues $\beta = \{\beta_1,\beta_2,...\beta_n \}$ and $\lambda = \{\lambda_1,\lambda_2,...\lambda_n \}$ in chosen order (say, decreasing).
Let $\forall \ell_i \in \ell, i\in\{1,2,...,n\},$
if $\sum_{i=1}^{n} (det(A-\ell_i I)-det(B-\ell_i I)) > 0$ then eigenvalues of $B$ are closer to $\ell$ and if $\sum_{i=1}^{n} (det(A-\ell_i I)-det(B-\ell_i I)) < 0$ then eigenvalues of $A$ are closer to $\ell$. Notice, since we don't actually need to calculate the eigenvalues of $A, B$, we don't need to solve the characteristic polynomial of $A, B$, It is enough to form the $det(A-\ell_i I)-det(B-\ell_i I)$, where $\ell_i$ is known then continue as described.
Is this solution right? and is it easier to find determinant of $n$ matrices than finding the eigenvalues of a $n \times n$ matrix?
