Consider square matrices $A$ and $B$, both of which are inverses of M-matrices, and a nonnegative diagonal matrix D. Is there a nice expression for the upper bound of $\rho \left( D^{-1} A D B \right)$ in terms of matrices $A$ and $B$? If $B=I$ then of course the spectral radius would be invariant to $D$, since $\rho \left( D^{-1} A D \right) = \rho(A)$, so we would have $\rho \left( D^{-1} A D B \right) = \rho (AB)$. How does this change when $B \neq I$?
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It seems there is a possibiilty of a bound only under some conditions. Let $D=diag(a,1,1,...,1)$ and let $a \rightarrow 0$ then first element of $D^{-1}$ tends to $\infty$. Then the first row of $D^{-1} A D B$ will be of the form: $[a_{11}, 1/a [a_{12},..,a_{1n}]]B$. Then for $||D^{-1}ADB||$ independent of $a$, we need, all eigen vectors of $B$ to be of the form: $[g_1, [g_2,...,g_n]]$ where $[g_2,...,g_n]$ is in null space of $[a_{12},....,a_{1n}]$. This need not happen in general case. If it doesnt happen then $||D^{-1}ADB||$ goes to $\infty$ due to $a \rightarrow 0$. So the bound may work only for cases where $||D^{-1}ADB||$ is far away from spectral radius $\rho(D^{-1}ADB)$. Use $A=[x,1;1,x]$ and $B=[0,1;1,0]$ and $D=[a,0;0,1]$. I think this example serves as a counter example to ur question and may not result in a bound independent of $a$.
Balaji sb
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The problem with this answer is that it would only apply to Hermitian matrices. Does it generalize beyond that case? – Andres May 06 '21 at 05:58
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Actually, even for the case in which $A$ and $B$ are Hermitian matrices, it is not clear to me how this works. Could you provide the detail? I know that if $A$ and $B$ commute then it is true that $\rho(AB) \leq \rho(A) \rho(B)$. But I don't see how this applies in this case. – Andres May 06 '21 at 20:15
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I have answered ur question on general case. Please see my edited answer. – Balaji sb May 07 '21 at 02:24
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Thanks! How does that apply to the case in which $B$ is the identity matrix? In that case, any vector is an eigenvector of $B$, but are they also in the null space of $[a_{12},...,a_{1n}]$? – Andres May 07 '21 at 04:31
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@Andres i have answered ur concern by editing the answer. – Balaji sb May 07 '21 at 05:00
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@Andres i gave u a clear example in the answer. Try it. – Balaji sb May 07 '21 at 05:14
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thanks this is very useful. Do you have any suggestions about how to do this in the case in which $B = A^T$? – Andres May 07 '21 at 17:24
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Let us continue this discussion in chat. – Andres May 07 '21 at 18:25
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@Andres Put $x=0$ in my example. In this case $B=A=A^T$. – Balaji sb May 08 '21 at 02:12
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Got it, thanks. Perhaps in my case things work the other way because the $A$ matrix is an inverse M-matrix. – Andres May 08 '21 at 05:00
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Consider two positive real square matrices $A$ and $B$, a positive vector $v$, and the positive diagonal matrix $D$ composed of the entries of $Av$.
Let $\rho(A)$ denote the spectral radius of $A$ and $j$ be the all ones column vector.
Consider the eigenvector $u$ of $A$ corresponding to $\rho(A)$ and let $D=\mbox{diag}(u)$.
Then $$BD^{-1}ADj=BD^{-1}(\rho(A)Dj)=\rho(A)Bj\leq \rho(A)r(B)j,$$ where $r(B)$ is the maximum row sum of $B$.
By Perron-Frobenius theorem, there exists a positive row vector $w^T$ which is the left eigenvector of $BD^{-1}AD$ corresponding to $\rho(BD^{-1}AD)$. Then $$\rho(BD^{-1}AD)w^Tj=w^TBD^{-1}ADj\leq \rho(A)r(B)w^Tj.$$ Since $w^Tj$ is a positive real number, we have $$\rho(BD^{-1}AD)\leq\rho(A)r(B).$$ Note that $\rho(D^{-1}ADB)=\rho(BD^{-1}AD)$. So $$\rho(D^{-1}ADB)\leq\rho(A)r(B).$$
Iris Wen
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u can swap $A$ and $B$ in ur answer and still get same bound ..so probably take $min(\rho(A) r(B),\rho(B)r(A))$ as upper bound. This can be proved by the fact $\rho(DBD^{-1}A)=\rho(D^{-1}ADB)$ – Balaji sb May 06 '21 at 02:41
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Thanks for this answer. Does it have to be the case that =() with being the eigenvector associated with $\rho(A)$? – Andres May 06 '21 at 06:09
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Perhaps we can use the fact that, under some conditions, any matrix $D$ can be written as a weighted average of the matrices $\text{diag}(v)$ with $v$ being the eigenvectors associated with the different eigenvalues of $A$, with those eigenvalues having lower absolute values than $\rho(A)$. – Andres May 06 '21 at 16:16
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One feature in my particular problem is that $B = A^T$, but I am not sure how to use this to establish a bound on $\rho(D^{-1} A D A^T)$ that only depends on $A$. – Andres May 07 '21 at 01:14