common eigenvector of a diagonal element cross-section

Denote by Mn⁢(𝒦) the set of all n×n matrices over 𝒦. Let di:Mn⁢(𝒦)⟶𝒦 be the function which extracts the ith diagonal element of a matrix, and let εi:𝒦n⟶𝒦 be the function which extracts the ith of a vector. Finally denote by [n] the set {1,…,n}.

Theorem 1.

Let K be a field. For any sequence A1,…,Ar∈Mn⁢(K) of upper triangular pairwise commuting matricesMathworldPlanetmath and every row index i∈[n], there exists u∈Kn∖{0} such that

Ak⁢𝐮=di⁢(Ak)⁢𝐮 for all k∈[r]. (1)

Let λk=di⁢(Ak) for all k∈[r], so that the problem is to find a common eigenvectorMathworldPlanetmathPlanetmathPlanetmath 𝐮 of A1,…,Ar whose corresponding eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath for Ak is λk. It is sufficient to find such a common eigenvector in the case that i=n is the least i∈[n] for which di⁢(Ak)=λk for all k∈[r], because if some smaller i also has this property then one can solve the corresponding problem for the i×i submatricesMathworldPlanetmath consisting of rows and columns 1 through i of A1,…,Ar, and then pad the common eigenvector of these submatrices with zeros to get a common eigenvector of the original A1,…,Ar.

By the existence of a characteristic matrix of a diagonal element cross-section ( there exists a matrix C in the unital algebra generated by A1,…,Ar such that di⁢(C)=1 if di⁢(Ak)=λk for all k∈[r], and di⁢(C)=0 otherwise; in other words that matrix C satisfies dn⁢(C)=1 and di⁢(C)=0 for all i<n. Since it is also upper triangular it follows that the matrix I-C has rank ( n-1, so the kernel of this matrix is one-dimensional. Let 𝐮=(u1,…,un)∈ker⁡(I-C) be such that un=1; it is easy to see that this is always possible (indeed, the only vector in this nullspaceMathworldPlanetmath with nth 0 is the zero vectorMathworldPlanetmath). This 𝐮 is the wanted eigenvector.

To see that it is an eigenvector of Ak, one may first observe that C commutes with this Ak, since the unital algebra of matrices to which C belongs is commutativePlanetmathPlanetmathPlanetmath ( This implies that Ak⁢𝐮∈ker⁡(I-C) since 𝟎=Ak⁢𝟎=Ak⁢(I-C)⁢𝐮=(I-C)⁢Ak⁢𝐮. As ker⁡(I-C) is one-dimensional it follows that Ak⁢𝐮=λ⁢𝐮 for some λ∈𝒦. Since Ak is upper triangular and un=1 this λ must furthermore satisfy λ=λ⁢un=εn⁢(λ⁢𝐮)=εn⁢(Ak⁢𝐮)=dn⁢(Ak)⁢un=dn⁢(Ak), 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