Proof of spectral radius bound $min_i sum_j a_{ij} le rho(A) le max_i sum_j a_{ij}$
I was reading one of the theorem in Roger A. Horn's Matrix Analysis and yet failed to understand how to prove it.
Let $A=[a_{ij}] in M_n$ be nonnegative and $rho(A)$ is spectral radius of $A$. Then
$$min_{1 le i le n} sum_{j=1}^n a_{ij} le rho(A) le max_{1 le i le n} sum_{j=1}^n a_{ij}$$
and
$$min_{1 le j le n} sum_{i=1}^n a_{ij} le rho(A) le max_{1 le j le n} sum_{i=1}^n a_{ij}$$
Can anyone help me to give me brief explaination and detailed proof of this theorem?
matrices eigenvalues-eigenvectors spectral-radius
edited Dec 1 at 10:58
glS
716520
asked Feb 26 at 8:51
Putri Alifiar
384
add a comment |
I was reading one of the theorem in Roger A. Horn's Matrix Analysis and yet failed to understand how to prove it.
Let $A=[a_{ij}] in M_n$ be nonnegative and $rho(A)$ is spectral radius of $A$. Then
$$min_{1 le i le n} sum_{j=1}^n a_{ij} le rho(A) le max_{1 le i le n} sum_{j=1}^n a_{ij}$$
and
$$min_{1 le j le n} sum_{i=1}^n a_{ij} le rho(A) le max_{1 le j le n} sum_{i=1}^n a_{ij}$$
Can anyone help me to give me brief explaination and detailed proof of this theorem?
matrices eigenvalues-eigenvectors spectral-radius
edited Dec 1 at 10:58
glS
716520
asked Feb 26 at 8:51
Putri Alifiar
384
add a comment |
I was reading one of the theorem in Roger A. Horn's Matrix Analysis and yet failed to understand how to prove it.
Let $A=[a_{ij}] in M_n$ be nonnegative and $rho(A)$ is spectral radius of $A$. Then
$$min_{1 le i le n} sum_{j=1}^n a_{ij} le rho(A) le max_{1 le i le n} sum_{j=1}^n a_{ij}$$
and
$$min_{1 le j le n} sum_{i=1}^n a_{ij} le rho(A) le max_{1 le j le n} sum_{i=1}^n a_{ij}$$
Can anyone help me to give me brief explaination and detailed proof of this theorem?
matrices eigenvalues-eigenvectors spectral-radius
edited Dec 1 at 10:58
glS
716520
asked Feb 26 at 8:51
Putri Alifiar
384
I was reading one of the theorem in Roger A. Horn's Matrix Analysis and yet failed to understand how to prove it.
Let $A=[a_{ij}] in M_n$ be nonnegative and $rho(A)$ is spectral radius of $A$. Then
$$min_{1 le i le n} sum_{j=1}^n a_{ij} le rho(A) le max_{1 le i le n} sum_{j=1}^n a_{ij}$$
and
$$min_{1 le j le n} sum_{i=1}^n a_{ij} le rho(A) le max_{1 le j le n} sum_{i=1}^n a_{ij}$$
Can anyone help me to give me brief explaination and detailed proof of this theorem?
matrices eigenvalues-eigenvectors spectral-radius
matrices eigenvalues-eigenvectors spectral-radius
edited Dec 1 at 10:58
glS
716520
asked Feb 26 at 8:51
Putri Alifiar
384
edited Dec 1 at 10:58
glS
716520
asked Feb 26 at 8:51
Putri Alifiar
384
edited Dec 1 at 10:58
glS
716520
edited Dec 1 at 10:58
glS
716520
edited Dec 1 at 10:58
glS
716520
716520
asked Feb 26 at 8:51
Putri Alifiar
384
asked Feb 26 at 8:51
Putri Alifiar
384
asked Feb 26 at 8:51
Putri Alifiar
384
384
add a comment |
add a comment |
1 Answer
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It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $rho(A)in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=rho(A)v$.
That is, for every $i$, $(*)$ $sum_ja_{i,j}v_j=rho(A) v_i$.
Let $v_k=sup_j v_j$. Then (take $i=k$ in $(*)$) $rho(A)leq sum_j a_{k,j}leq max_i sum_j a_{i,j}$ .
Let $v_l=inf_j v_j$. Then (take $i=l$ in $(*)$) $rho(A)geq sum_j a_{l,j}geq min_i sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.
answered Dec 2 at 21:29
loup blanc
22.5k21750
add a comment |
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It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $rho(A)in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=rho(A)v$.
That is, for every $i$, $(*)$ $sum_ja_{i,j}v_j=rho(A) v_i$.
Let $v_k=sup_j v_j$. Then (take $i=k$ in $(*)$) $rho(A)leq sum_j a_{k,j}leq max_i sum_j a_{i,j}$ .
Let $v_l=inf_j v_j$. Then (take $i=l$ in $(*)$) $rho(A)geq sum_j a_{l,j}geq min_i sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.
answered Dec 2 at 21:29
loup blanc
22.5k21750
add a comment |
It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $rho(A)in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=rho(A)v$.
That is, for every $i$, $(*)$ $sum_ja_{i,j}v_j=rho(A) v_i$.
Let $v_k=sup_j v_j$. Then (take $i=k$ in $(*)$) $rho(A)leq sum_j a_{k,j}leq max_i sum_j a_{i,j}$ .
Let $v_l=inf_j v_j$. Then (take $i=l$ in $(*)$) $rho(A)geq sum_j a_{l,j}geq min_i sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.
answered Dec 2 at 21:29
loup blanc
22.5k21750
add a comment |
It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $rho(A)in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=rho(A)v$.
That is, for every $i$, $(*)$ $sum_ja_{i,j}v_j=rho(A) v_i$.
Let $v_k=sup_j v_j$. Then (take $i=k$ in $(*)$) $rho(A)leq sum_j a_{k,j}leq max_i sum_j a_{i,j}$ .
Let $v_l=inf_j v_j$. Then (take $i=l$ in $(*)$) $rho(A)geq sum_j a_{l,j}geq min_i sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.
answered Dec 2 at 21:29
loup blanc
22.5k21750
It suffices to show the result when $A$ is a positive matrix (the $(a_{i,j})$ are $>0$); indeed we conclude, using the continuity of $rho(A)$ wrt the $(a_{i,j})$.
When $A$ is positive, $rho(A)in spectrum(A)$ and $A$ admits an eigenvector $v=(v_i)$ s.t. $v_i>0$ and $Av=rho(A)v$.
That is, for every $i$, $(*)$ $sum_ja_{i,j}v_j=rho(A) v_i$.
Let $v_k=sup_j v_j$. Then (take $i=k$ in $(*)$) $rho(A)leq sum_j a_{k,j}leq max_i sum_j a_{i,j}$ .
Let $v_l=inf_j v_j$. Then (take $i=l$ in $(*)$) $rho(A)geq sum_j a_{l,j}geq min_i sum_j a_{i,j}$.
For the last two inequalities, consider $A^T$. Note that $A^T$ is positive and has same spectral radius as $A$.
answered Dec 2 at 21:29
loup blanc
22.5k21750
answered Dec 2 at 21:29
loup blanc
22.5k21750
answered Dec 2 at 21:29
loup blanc
22.5k21750
answered Dec 2 at 21:29
loup blanc
22.5k21750
22.5k21750
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add a comment |
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