proof of theorem for normal matrices

1) (AH=g(A)A is normal)

Keeping in mind that every matrix commutes with its own powers, let’s compute


which shows A to be normal.

2) (A is normal AH=g(A))

Let λ1,λ2,,λr , 1rn be the distinct eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath of A, and let Λ = diag{λ1,λ2,,λn}; then it’s possible to find a (r-1)-degree polynomial g(t) such that g(λi)=λi 1ir, solving the r×r linear Vandermonde system:


Since these r eigenvalues are distinct, the Vandermonde matrixMathworldPlanetmath is full rank, and the linear system admits a unique solution; so a (r-1)-degree polynomial g(t) can be found such that g(λi)=λi 1ir and therefore g(λi)=λi 1in. Writing these equations in matrix form, we have


By Schur’s decomposition theorem, a unitary matrixMathworldPlanetmath U and an upper triangular matrixMathworldPlanetmath T exist such that


and since A is normal we have T=Λ.

Let’s evaluate g(A).


But, keeping in mind that UHU=I,


and so

g(A) = i=0r-1ai(UΛiUH)
= U(i=0r-1aiΛi)UH
= Ug(Λ)UH

which is the thesis.

Remark: note that this is a constructive proofMathworldPlanetmath, giving explicitly a way to find g(t) polynomial by solving Vandermonde system in the eigenvalues.


Let A=12[1+j-1-j1+j1+j] (which is easily checked to be normal),

with  U =12[1-jj-1]. Then σ(A)={1,j} and the Vandermonde system is


from which we find


A simple calculation yields

Title proof of theorem for normal matrices
Canonical name ProofOfTheoremForNormalMatrices
Date of creation 2013-03-22 15:36:36
Last modified on 2013-03-22 15:36:36
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 16
Author Andrea Ambrosio (7332)
Entry type Proof
Classification msc 15A21