Linear Algebra_Spectral radius of matrix 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. 
 A: 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.
A: 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.
