Talk:Hessenberg matrix

Latest comment: 15 years ago by Jitse Niesen in topic Counterexample

Counterexample edit

What exactly is the counterexample to the statement,

"if A is upper Hesesnberg and T is upper triangular, then AT and TA are upper Hessenberg."

The proof is rather simple... and I don't see where the counteexample might arise. Here is my proof.

Suppose T is upper triangular and AT is upper Hessenberg.

Case 1: Suppose T is invertible. Then, T(AT) is upper Hessenberg (from prev. proof) and T^-1 is upper triangular (from prev proof). Similarly, (TAT)T^-1= TA is upper Hessenberg (from prev. proof). --> this part is unquestionable.

Case 2: Suppose T is not invertible. This means there exists one or more element on the diagonal of T that equals 0 (from the fact that det(T)=product(det(T_ii),for all i). However, there exists an invertible sequence of matrices arbitarily close to T such that lim_k->infinity (T_k)=T. T_k has non-zero diagonal elements but has one or more elements that are arbitarily close to 0. Clearly, T_k is invertible. Using similar logic as in case 1, lim_k->infinity (TAT)T_k^-1=TA is an upper Hessenberg matrix. Since T_k is arbitarrily close to T and for all invertible T_k, TA is upper Hessenberg matrix, we conclude that TA must be a Hessenberg matrix. —Preceding unsigned comment added by Sjayzzang (talkcontribs) 2009-04-15T20:11:16Z

The counterexample was to a different statement: If T is upper triangular and AT is upper Hessenberg, then TA is upper Hessenberg. Counterexamples are actually easy to find: for any integer n ≥ 3, let A=En,1 and T=En,n. Then T is upper triangular (diagonal even), AT=0 is upper Hessenberg, but TA = A is not upper Hessenberg. JackSchmidt (talk) 21:59, 15 April 2009 (UTC)Reply
I think Sjayzzang's attempted proof is indeed for that statement. The weak point seems to be that lim_k->infinity (TAT)T_k^-1 is not necessarily TA. -- Jitse Niesen (talk) 01:20, 16 April 2009 (UTC)Reply