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a) Show that the largest eigenvalue of a non-negative matrix is upper bounded by its largest row sum.

b) For a non-negative matrix $M$, show that the largest eigenvalue of $M$ is upper bounded by the largest row sum of $DMD^{-1}$, for any positive diagonal matrix $D$.

I've tried multiplying by one of the eigenvalues and making comparisons, but i couldn't find any that is true only for the largest row sum. Also, i've tried multiplying the matrix by the only 1's vector, since that would result on the vector with the row sums, but i didn't find any useful inequalities. The matrix may not be symmetric, so we can't assume it has a basis of eigenvectors.

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(a) Let $\lambda$ be an eigen value of the non-negative matrix $A.$ Then $Ax=\lambda x \implies \lambda |x_i|= |\sum_{j=1}^n a_{ij}x_j|\leq (\sum_{j=1}^n a_{ij})\max_j|x_j|\implies \lambda \max_i |x_i|\leq \max_i (\sum_{j=1}^n a_{ij})\max_j|x_j|\implies \lambda\leq\max_i (\sum_{j=1}^n a_{ij})$

(b) Again $\lambda$ be an eigen value of the non-negative matrix $A.$ Then $Ax=\lambda x \implies AD^{-1}Dx=\lambda D^{-1}Dx\implies DAD^{-1}Dx=\lambda DD^{-1}Dx\implies DAD^{-1}y=\lambda y$ for $y=Dx.$ Same proof as (a) follows because $DAD^{-1}$ is non-negative for $D$ positive diagonal.

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