On the supremum norm of matrices
$begingroup$
Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that
$||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.
Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?
I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.
Please help
NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$
linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets
asked Jan 11 at 6:04
user521337user521337
1,2201417
$endgroup$
add a comment |
$begingroup$
Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that
$||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.
Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?
I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.
Please help
NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$
linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets
asked Jan 11 at 6:04
user521337user521337
1,2201417
$endgroup$
add a comment |
$begingroup$
Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that
$||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.
Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?
I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.
Please help
NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$
linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets
asked Jan 11 at 6:04
user521337user521337
1,2201417
$endgroup$
Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that
$||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.
Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?
I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.
Please help
NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$
linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets
linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets
asked Jan 11 at 6:04
user521337user521337
1,2201417
asked Jan 11 at 6:04
user521337user521337
1,2201417
edited Jan 11 at 6:26
user521337
asked Jan 11 at 6:04
user521337user521337
1,2201417
asked Jan 11 at 6:04
user521337user521337
1,2201417
asked Jan 11 at 6:04
user521337user521337
1,2201417
1,2201417
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.
We note that
$$
|E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
$$
Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.
Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.
We note that
$$
|E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
$$
Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.
Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
add a comment |
$begingroup$
Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.
We note that
$$
|E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
$$
Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.
Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
add a comment |
$begingroup$
Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.
We note that
$$
|E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
$$
Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.
Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
$endgroup$
Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.
We note that
$$
|E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
$$
Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.
Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
edited Jan 11 at 14:56
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
answered Jan 11 at 6:37
OmnomnomnomOmnomnomnom
129k794188
129k794188
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
add a comment |
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
why are we try to show that the disks will not overlap ?
$endgroup$
– user521337
Jan 11 at 8:31
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Because if the disks have no overlap, then each contains one eigenvalue.
$endgroup$
– Omnomnomnom
Jan 11 at 14:55
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
$begingroup$
Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
$endgroup$
– Omnomnomnom
Jan 11 at 17:03
add a comment |
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