characteristic matrix of diagonal element cross-section

Denote by $\mathrm{M}_{n}(\mathcal{K})$ the set of all $n\times n$ matrices over $\mathcal{K}$ . Let $d_{i}\colon\mathrm{M}_{n}(\mathcal{K})\longrightarrow\mathcal{K}$ be the function which extracts the $i$ th diagonal element of a matrix. Finally denote by $[n]$ the set $\{1,\ldots,n\}$ .

Lemma.

Let $\mathcal{K}$ be a field. Let a sequence $A_{1},\ldots,A_{r}\in\mathrm{M}_{n}(\mathcal{K})$ of upper triangular matrices be given, and denote by $\mathcal{A}\subseteq\mathrm{M}_{n}(\mathcal{K})$ the unital algebra generated by these matrices. For every sequence $\lambda_{1},\ldots,\lambda_{r}\in\mathcal{K}$ of scalars there exists a matrix $C\in\mathcal{A}$ such that

d_{i}(C)=\begin{cases}1&\text{if \(d_{i}(A_{k})=\lambda_{k}\) for all \(k\in[r% ]\),}\\ 0&\text{otherwise}\end{cases}

for all $i\in[n]$ .

A diagonal element of an upper triangular matrix is of course an eigenvalue of that matrix, so the particular $\lambda_{1},\ldots,\lambda_{r}$ that one plugs into this lemma is typically either a sequence of eigenvalues for the given matrices, or a sequence of values that one 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 $\mathcal{A}$ than what was explicitly required above. A particular example is that if the given matrices commute then $\mathcal{A}$ will be a commutative (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=\bigl{\{}i\in[n]\,\bigm{|}\,\text{\(d_{i}(A_{k})\neq\lambda_{k}\) for some % \(k\in[r]\)}\bigr{\}}\text{,}

and for each $i\in S$ let $k_{i}\in[r]$ be such that $d_{i}(A_{k_{i}})\neq\lambda_{k_{i}}$ . Define

B_{i}=\begin{cases}A_{k_{i}}-d_{i}(A_{k_{i}})I&\text{if \(i\in S\),}\\ I&\text{otherwise}\end{cases}

for all $i\in[n]$ and let $B=\prod_{i=1}^{n}B_{i}$ . Since all the matrices involved are upper triangular, a diagonal element in $B$ is simply the product of the corresponding diagonal elements in all of $B_{1},\ldots,B_{n}$ . If $i\in S$ then $d_{i}(B_{i})=d_{i}\bigl{(}A_{k_{i}}-\nobreak d_{i}(A_{k_{i}})I\bigr{)}=d_{i}(A% _{k_{i}})-d_{i}(A_{k_{i}})=0$ and thus $d_{i}(B)=0$ for $i\in S$ . If instead $i\in[n]\setminus S$ then

\displaystyle d_{i}(B)=\prod_{j=1}^{n}d_{i}(B_{i})=\prod_{j\in S}d_{i}\bigl{(}% A_{k_{j}}-d_{j}(A_{k_{j}})I\bigr{)}=\\ \displaystyle=\prod_{j\in S}\bigl{(}d_{i}(A_{k_{j}})-d_{j}(A_{k_{j}})\bigr{)}=% \prod_{j\in S}\bigl{(}\lambda_{k_{j}}-d_{j}(A_{k_{j}})\bigr{)}=:b\text{.}

This $b$ is by definition of $k_{j}$ nonzero, and since it is independent of $i$ it follows that it is the only nonzero value of a diagonal element of $B$ . Hence the wanted $C=b^{-1}B$ . ∎

Title	characteristic matrix of diagonal element cross-section
Canonical name	CharacteristicMatrixOfDiagonalElementCrosssection
Date of creation	2013-03-22 15:29:45
Last modified on	2013-03-22 15:29:45
Owner	lars_h (9802)
Last modified by	lars_h (9802)
Numerical id	4
Author	lars_h (9802)
Entry type	Theorem
Classification	msc 15A30