# 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 ${d}_{i}:{\mathrm{M}}_{n}(\mathcal{K})\u27f6\mathcal{K}$ be the function which extracts the $i$th diagonal element of a matrix, and let ${\epsilon}_{i}:{\mathcal{K}}^{n}\u27f6\mathcal{K}$ be the function which extracts the $i$th of a vector. Finally denote by $[n]$ the set $\{1,\mathrm{\dots},n\}$.

###### Theorem 1.

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

$${A}_{k}\mathbf{u}={d}_{i}({A}_{k})\mathbf{u}\mathit{\hspace{1em}}\mathit{\text{for all}}k\in [r]\mathit{\text{.}}$$ | (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 ${A}_{1},\mathrm{\dots},{A}_{r}$
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 ${A}_{1},\mathrm{\dots},{A}_{r}$, and then pad the common eigenvector
of these submatrices with zeros to get a common eigenvector of the
original ${A}_{1},\mathrm{\dots},{A}_{r}$.

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 ${A}_{1},\mathrm{\dots},{A}_{r}$ 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 $$. 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 $\mathbf{u}=({u}_{1},\mathrm{\dots},{u}_{n})\in \mathrm{ker}(I-C)$ 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
${A}_{k}\mathbf{u}\in \mathrm{ker}(I-C)$ since $\mathrm{\U0001d7ce}={A}_{k}\mathrm{\U0001d7ce}={A}_{k}(I-C)\mathbf{u}=(I-C){A}_{k}\mathbf{u}$. As $\mathrm{ker}(I-C)$ 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
$\lambda =\lambda {u}_{n}={\epsilon}_{n}(\lambda \mathbf{u})={\epsilon}_{n}({A}_{k}\mathbf{u})={d}_{n}({A}_{k}){u}_{n}={d}_{n}({A}_{k})$, which is
indeed what the eigenvalue was claimed to be.
∎

Title | common eigenvector of a diagonal element cross-section |
---|---|

Canonical name | CommonEigenvectorOfADiagonalElementCrosssection |

Date of creation | 2013-03-22 15:30:41 |

Last modified on | 2013-03-22 15:30:41 |

Owner | lars_h (9802) |

Last modified by | lars_h (9802) |

Numerical id | 4 |

Author | lars_h (9802) |

Entry type | Theorem |

Classification | msc 15A18 |

Related topic | CommutingMatrices |