Prove that a small shift in the diagonal term leads to smaller spectral radius (for Perron-Frobenius theorem) On Wikipedia, the proof for Perron Frobenius theorem in the strictly positive case has a confusing step: Suppose $T=A^m-\epsilon I$, where $\epsilon$ is smaller than the smallest diagonal term of strictly positive matrix $A^m$. Then it straightforwardly claim that $\rho(T)\le\rho(A^m)$. How can one prove this step?
Eigenvalues of the matrix $A^m$ are also the roots of the polynomial equation
$$\det(A^m-\lambda I)=0$$
Thus $\lambda$ (roots of the above equation) are also eigenvalues of $A^m$. Now, note that
$$\det(A^m-\lambda I) = \det (\,\,(A^m-\epsilon I)-(\lambda-\epsilon) I\,\,)=\det (T-(\lambda - \epsilon)I)$$
Thus, following earlier argument, $\lambda - \epsilon$ should be the eigenvalues of $T$. Now, note that $$\rho(A^m)=\max_{i}|\lambda_i|$$
$$\rho(T)=\max_{i}|\lambda_i-\epsilon|$$
Now, note that 
$$\lambda_1 + \dots + \lambda_n = T_{11}+\dots + T_{nn}$$ and also $T_{ii} &gt; 0$. Let $T_{nn}$ be the smallest diagonal entry (w.l.o.g). This implies that $$\rho(A^m)= \max_{i}|\lambda_i|\geq \max_{i}|\lambda_i| - T_{nn}\geq \max_{i}|\lambda_i| - \epsilon \geq \rho(T)$$
Eigenvalues of the matrix $A^m$ are also the roots of the polynomial equation
$$\det(A^m-\lambda I)=0$$
Thus $\lambda$ (roots of the above equation) are also eigenvalues of $A^m$. Now, note that
$$\det(A^m-\lambda I) = \det (\,\,(A^m-\epsilon I)-(\lambda-\epsilon) I\,\,)=\det (T-(\lambda - \epsilon)I)$$
Thus, following earlier argument, $\lambda - \epsilon$ should be the eigenvalues of $T$. Now, note that $$\rho(A^m)=\max_{i}|\lambda_i|$$
$$\rho(T)=\max_{i}|\lambda_i-\epsilon|$$
Now, note that 
$$\lambda_1 + \dots + \lambda_n = T_{11}+\dots + T_{nn}$$ and also $T_{ii} &gt; 0$. Let $T_{nn}$ be the smallest diagonal entry (w.l.o.g). This implies that $$\rho(A^m)= \max_{i}|\lambda_i|\geq \max_{i}|\lambda_i| - T_{nn}\geq \max_{i}|\lambda_i| - \epsilon \geq \rho(T)$$