# common eigenvector of a 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, and let $\varepsilon_{i}\colon\mathcal{K}^{n}\longrightarrow\mathcal{K}$ be the function which extracts the $i$th of a vector. Finally denote by $[n]$ the set $\{1,\ldots,n\}$.

###### Theorem 1.

Let $\mathcal{K}$ be a field. For any sequence $A_{1},\ldots,A_{r}\in\mathrm{M}_{n}(\mathcal{K})$ of upper triangular pairwise commuting matrices and every row index $i\in[n]$, there exists $\mathbf{u}\in\mathcal{K}^{n}\setminus\{0\}$ such that

 $A_{k}\mathbf{u}=d_{i}(A_{k})\mathbf{u}\quad\text{for all $$k\in[r]$$.}$ (1)
###### Proof.

Let $\lambda_{k}=d_{i}(A_{k})$ for all $k\in[r]$, so that the problem is to find a common eigenvector $\mathbf{u}$ of $A_{1},\ldots,A_{r}$ whose corresponding eigenvalue for $A_{k}$ is $\lambda_{k}$. It is sufficient to find such a common eigenvector in the case that $i=n$ is the least $i\in[n]$ for which $d_{i}(A_{k})=\lambda_{k}$ for all $k\in[r]$, because if some smaller $i$ also has this property then one can solve the corresponding problem for the $i\times i$ submatrices consisting of rows and columns $1$ through $i$ of $A_{1},\ldots,A_{r}$, and then pad the common eigenvector of these submatrices with zeros to get a common eigenvector of the original $A_{1},\ldots,A_{r}$.

By the existence of a characteristic matrix of a diagonal element cross-section (http://planetmath.org/CharacteristicMatrixOfDiagonalElementCrossSection) there exists a matrix $C$ in the unital algebra generated by $A_{1},\ldots,A_{r}$ such that $d_{i}(C)=1$ if $d_{i}(A_{k})=\lambda_{k}$ for all $k\in[r]$, and $d_{i}(C)=0$ otherwise; in other words that matrix $C$ satisfies $d_{n}(C)=1$ and $d_{i}(C)=0$ for all $i. Since it is also upper triangular it follows that the matrix $I-C$ has rank (http://planetmath.org/RankLinearMapping) $n-1$, so the kernel of this matrix is one-dimensional. Let $\mathbf{u}=(u_{1},\ldots,u_{n})\in\ker(I-\nobreak C)$ be such that $u_{n}=1$; it is easy to see that this is always possible (indeed, the only vector in this nullspace with $n$th $0$ is the zero vector). This $\mathbf{u}$ is the wanted eigenvector.

To see that it is an eigenvector of $A_{k}$, one may first observe that $C$ commutes with this $A_{k}$, since the unital algebra of matrices to which $C$ belongs is commutative (http://planetmath.org/Commutative). This implies that $A_{k}\mathbf{u}\in\ker(I-\nobreak C)$ since $\mathbf{0}=A_{k}\mathbf{0}=A_{k}(I-\nobreak C)\mathbf{u}=(I-\nobreak C)A_{k}% \mathbf{u}$. As $\ker(I-\nobreak C)$ is one-dimensional it follows that $A_{k}\mathbf{u}=\lambda\mathbf{u}$ for some $\lambda\in\mathcal{K}$. Since $A_{k}$ is upper triangular and $u_{n}=1$ this $\lambda$ must furthermore satisfy $\lambda=\lambda u_{n}=\varepsilon_{n}(\lambda\mathbf{u})=\varepsilon_{n}(A_{k}% \mathbf{u})=d_{n}(A_{k})u_{n}=d_{n}(A_{k})$, which is indeed what the eigenvalue was claimed to be. ∎

Title common eigenvector of a diagonal element cross-section CommonEigenvectorOfADiagonalElementCrosssection 2013-03-22 15:30:41 2013-03-22 15:30:41 lars_h (9802) lars_h (9802) 4 lars_h (9802) Theorem msc 15A18 CommutingMatrices