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We assume the spectral radius of the matrix T is less(not equal) than one. Also, the matrix T is nonsingular,i.e. the spectral radius of the matrix T is bigger(not equal) than zero. How we can show that the spectral radius of the matrix T1 is less(not equal) than one, where T1=D*|T|, D=diag(d_11, d_22,...,d_nn) and d_ii<1, i=1,2,...,n. D is the n*n diagonal matrix with diagonal entries d_ii and |T| is the absolute value of the matrix T.

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  • $\begingroup$ Welcome to MSE. Please use formatting tools to present your questions in a readable form, lest they will be ignored. $\endgroup$ – uniquesolution Aug 31 '17 at 12:45
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Well, you can calculate the eigenvalues of $T_1$ directly. Since the spectral radius of $T$ is positive and less than 1, i.e. $$\max\{|\lambda_1|,\cdots,|\lambda_n|\}\le 1$$ where $\lambda_i, i=1,\cdots,n$ are all the eigenvalues of T. Furthermore, because T is nonsingular, we can conclude that $$0<|\lambda_i|<1, i=1,\cdots,n$$ $$|T|= \lambda_1\cdots\lambda_n$$ We let $a = |T|$.Thus $$0< |a|=|\lambda_1|\cdots|\lambda_n|<1$$ Now, let us calculate the eigenvalues of $T_1$; we denote these eigenvalues by $\delta_1,\cdots,\delta_n$. Since $T_1 = D*|T|$, so we have $$\delta_i = d_{ii}*a,i=1,\cdots,n$$ So, $|\delta_i|=|d_{ii}||a|\le |d_{ii}|,i=1,\cdots,n$ (we put $\le$ here to consider into the case that maybe $d_{jj}=0$ for some $1\le j\le n$). So there exists a problem, as you could see, that if you don't provide $|d_{ii}|<1,i=1,\cdots,n$ in the question, we actually can not come to the conclusion that the spectral radius of $T_1$ is less than 1. Maybe you missing something as follow: $0<d_{ii}<1,i=1,\cdots,n$, Then we can prove the problem.

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You cannot prove that, because it isn't true in general. If it were true, then in the limiting case we would have $\rho(|T|)=\|T\|_2\le1$ whenever $\rho(T)\le1$. Yet this is clearly false when $T$ is the sum of $I$ plus a sufficiently large strictly upper triangular part.

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