$(I-T)x=d$ always has non-negative solution $x$ if $d$ non-negative. Why $\rho(T)<1$?

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This question comes from a model in economy developed by Leontief.

Definition If for every non-negative vector $d$, there is a non-negative vector $x$ satisfying $(I-T)x=d$, we say $T$ is feasible.

If for every non-negative vector $v$, there is a non-negative vector $p$ satisfying $(I-T)^{\mathrm{T}}p=v$, we say $T$ is profitable.

Problem Prove that if $T$ is feasible (or profitable), then $\rho(T)<1$.

Here $\rho(T)$ stands for spectual radius, i.e., the maximal norm of eigenvalues of $T$.

I wonder where to get start with. Any help would be appreciated!