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$$A = \begin{pmatrix} A_1 & A_2 \\ A_3 & A_4 \end{pmatrix}$$

is a block matrix with all blocks of order $n$ by $n$. Blocks $A_2$ and $A_3$ are diagonal matrices with non-negative entries. Blocks $A_1$ and $A_4$ are tridiagonal matrices and also monotone. Prove that $A$ is a monotone matrix. I have approached the question as : Writing $$A = \begin{pmatrix} A_1 & 0 \\ 0 & A_4 \end{pmatrix} + \begin{pmatrix} 0 & A_2 \\ A_3 & 0 \end{pmatrix} = A1 + A2 $$ We have $A1$ is monotone matrix because it is a block diagonal matrix whose diagonals are monotone matrices. Now how do I prove that adding $A2$ to $A1$ will also give a monotone matrix ($A_2$ and $A_3$ are diagonal matrices.)
(A is monotone if for every column vector v, $Av≥0⟹v=0$. Equivalently, it means that $A^{−1}$ exists and has only non-negative entries.)

founderv
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1 Answers1

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Does not seem to be true in general:

See: Look at Inverse of a $2 \times 2$ block matrix

For $A^{-1}$ has non-negative entries $ \implies (A_4-A_3A_1^{-1}A_2)^{-1}$ has non-negative and $-(A_4-A_3A_1^{-1}A_2)^{-1} A_3 A_1^{-1}$ has non-negative entries.

Since $A_3$,$A_1^{-1}$ and $(A_4-A_3A_1^{-1}A_2)^{-1}$ has non-negative entries, implies $-(A_4-A_3A_1^{-1}A_2)^{-1} A_3 A_1^{-1}$ is not a matrix with positive entries.

Implies $A_3 = 0$.

By similar argument, we can conclude that $A_2 = 0$ and the resulting matrix is block diagonal monotone matrix and hence monotone.

Hence $A$ is monotone $\iff$ $A_2 = 0, A_3 = 0$.