characteristic matrix of diagonal element cross-section

Denote by ${\mathrm{M}}_n(𝒦)$ the set of all $n\times n$ matrices over$𝒦$. Let $d_i:{\mathrm{M}}_n(𝒦)⟶𝒦$ bethe function which extracts the $i$th diagonal element of a matrix.Finally denote by $[n]$ the set $\{1,\mathrm{\dots },n\}$.

Lemma.

Let $\mathrm{K}$ be a field.Let a sequence $A_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{A}_r\mathrm{\in }{\mathrm{M}}_n\mathit{}\mathrm{(}\mathrm{K}\mathrm{)}$ of uppertriangular matrices be given, and denote by $\mathrm{A}\mathrm{\subseteq }{\mathrm{M}}_n\mathit{}\mathrm{(}\mathrm{K}\mathrm{)}$ the unital algebra generated by these matrices.For every sequence ${\lambda }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\lambda }_r\mathrm{\in }\mathrm{K}$ ofscalars there exists a matrix $C\mathrm{\in }\mathrm{A}$ such that

for all $i\mathrm{\in }\mathrm{[}n\mathrm{]}$.

A diagonal element of an upper triangular matrix is of course aneigenvalue of that matrix, so the particular ${\lambda }_1,\mathrm{\dots },{\lambda }_r$that one plugs into this lemma is typically either a sequence ofeigenvalues for the given matrices, or a sequence of values thatone thinks may be eigenvalues for these matrices.The “cross-section” in the title refers to that there is one${\lambda }_k$ for each matrix $A_k$.

The result gets more interesting if one knows something more about$𝒜$ than what was explicitly required above. A particular exampleis that if the given matrices commute then $𝒜$ will be acommutative (http://planetmath.org/Commutative) algebra and consequently$C$ will commute with all of the given matrices as well.

Proof.

Let $S$ be the set of those row indices $i$ for which $d_i(C)$should be $0$, i.e.,

$$S=\left\{i\in [n]\right|d_i(A_k)\ne {\lambda }_k\text{ for some }k\in [r]\}\text{,}$$

and for each $i\in S$ let $k_i\in [r]$ be such that$d_i(A_{k_i})\ne {\lambda }_{k_i}$. Define

for all $i\in [n]$ and let $B={\prod }_{i=1}^n{B}_i$. Since allthe matrices involved are upper triangular, a diagonal element in $B$is simply the product of the corresponding diagonal elements in allof $B_1,\mathrm{\dots },B_n$. If $i\in S$ then $d_i(B_i)=d_i\left(A_{k_i}-d_i(A_{k_i})I\right)=d_i(A_{k_i})-d_i(A_{k_i})=0$ and thus $d_i(B)=0$ for $i\in S$. If instead $i\in [n]\setminus S$ then

$$\begin{array}{c}{d}_i(B)=\prod _{j=1}^n{d}_i(B_i)=\prod _{j\in S}{d}_i\left(A_{k_j}-d_j(A_{k_j})I\right)=\hfill \\ \hfill =\prod _{j\in S}(d_i(A_{k_j})-d_j(A_{k_j}))=\prod _{j\in S}({\lambda }_{k_j}-d_j(A_{k_j}))=:b\text{.}\end{array}$$

This $b$ is by definition of $k_j$ nonzero, and since it isindependent of $i$ it follows that it is the only nonzero value of adiagonal element of $B$. Hence the wanted $C=b^{-1}B$.∎