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Let $D = \operatorname{diag}(d_1, d_2, \ldots, d_n)$ with each of its diagnal entry $d_i \in (0,1)$. $B$ is a non-negative matrix. Consider the following matrix $$ A = DI + (I-D)B, $$ where $I$ is the identity matrix. Each row of $A$ is a linear combination of the corresponding rows of $I$ and $B$.

My question is: Is there any relationship between the spectral radius of these two matrices?

Special Case

If $D = tI$ with $t\in(0,1)$, then for any eigenvalue $\lambda$ of $B$, $t+(1-t)\lambda$ is the eigenvalue of $A$. Hence, we have $$ \rho(A) = t + (1-t)\rho(B), $$ where $\rho(\cdot)$ denotes the spectral radius of the given matrix. This shows $\rho(A)$ lies in the open interval between $\rho(B)$ and $1$, and $$ \rho(A) < 1 \Leftrightarrow \rho(B)<1. $$ For general $D$, can we expect a similar conclusion holds?

Any thoughts on this problem are appreciated.

maphado fan
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  • If $B$ is an irreducible non-negative matrix, let $v>0$ be the eigenvector associated with $\rho(B)$. If $\rho(B)<1$, then one has $$ \rho(B) v < \rho(B) v + (1-\rho(B)) Dv = A v < \rho(B)v+(1-\rho(B))v = v. $$ This shows that $\rho(B) < \rho(A) < 1$. – maphado fan Nov 15 '23 at 06:05

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