Mixing
If $leftVert ArightVert geq c$ then $left|lambdaright|>c$ for all eigenvalues of $A$
$begingroup$
Let $Aintext{Mat}_{ntimes n}$ be an invertible matrix such that$leftVert ArightVert geq c$
where $leftVert cdotrightVert $ is the matrix norm . Is it true
that for every $lambda$ eigenvalue of $A$ we have $left|lambdaright|geq c$?
I know that $leftVert ArightVert geqleft|lambdaright|$ but
why would $left|lambdaright|geq c$?
Also Im familiar with the norm inequality $left|Axright|leqleftVert ArightVert left|xright|$
but couldn't use it in favor. Any suggestions?
linear-algebra matrices norm
asked Jan 18 at 22:40
JonJon
565414
$endgroup$
add a comment |
$begingroup$
Let $Aintext{Mat}_{ntimes n}$ be an invertible matrix such that$leftVert ArightVert geq c$
where $leftVert cdotrightVert $ is the matrix norm . Is it true
that for every $lambda$ eigenvalue of $A$ we have $left|lambdaright|geq c$?
I know that $leftVert ArightVert geqleft|lambdaright|$ but
why would $left|lambdaright|geq c$?
Also Im familiar with the norm inequality $left|Axright|leqleftVert ArightVert left|xright|$
but couldn't use it in favor. Any suggestions?
linear-algebra matrices norm
asked Jan 18 at 22:40
JonJon
565414
$endgroup$
1
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44
add a comment |
$begingroup$
Let $Aintext{Mat}_{ntimes n}$ be an invertible matrix such that$leftVert ArightVert geq c$
where $leftVert cdotrightVert $ is the matrix norm . Is it true
that for every $lambda$ eigenvalue of $A$ we have $left|lambdaright|geq c$?
I know that $leftVert ArightVert geqleft|lambdaright|$ but
why would $left|lambdaright|geq c$?
Also Im familiar with the norm inequality $left|Axright|leqleftVert ArightVert left|xright|$
but couldn't use it in favor. Any suggestions?
linear-algebra matrices norm
asked Jan 18 at 22:40
JonJon
565414
$endgroup$
Let $Aintext{Mat}_{ntimes n}$ be an invertible matrix such that$leftVert ArightVert geq c$
where $leftVert cdotrightVert $ is the matrix norm . Is it true
that for every $lambda$ eigenvalue of $A$ we have $left|lambdaright|geq c$?
I know that $leftVert ArightVert geqleft|lambdaright|$ but
why would $left|lambdaright|geq c$?
Also Im familiar with the norm inequality $left|Axright|leqleftVert ArightVert left|xright|$
but couldn't use it in favor. Any suggestions?
linear-algebra matrices norm
linear-algebra matrices norm
asked Jan 18 at 22:40
JonJon
565414
asked Jan 18 at 22:40
JonJon
565414
asked Jan 18 at 22:40
JonJon
565414
asked Jan 18 at 22:40
JonJon
565414
asked Jan 18 at 22:40
JonJon
565414
565414
1
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44
add a comment |
1
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44
1
1
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
No. Consider any diagonal matrix of the form $A = begin{pmatrix} c & 0 \ 0 & varepsilon I_{n-1} end{pmatrix}$ where $I_{n-1}$ is the $n-1 times n-1$ identity. Then $||A|| = c$, so clearly $||A|| geq c$, but all its other eigenvalues are $varepsilon < c$.
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
$endgroup$
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
|
show 1 more comment
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
No. Consider any diagonal matrix of the form $A = begin{pmatrix} c & 0 \ 0 & varepsilon I_{n-1} end{pmatrix}$ where $I_{n-1}$ is the $n-1 times n-1$ identity. Then $||A|| = c$, so clearly $||A|| geq c$, but all its other eigenvalues are $varepsilon < c$.
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
$endgroup$
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
|
show 1 more comment
$begingroup$
No. Consider any diagonal matrix of the form $A = begin{pmatrix} c & 0 \ 0 & varepsilon I_{n-1} end{pmatrix}$ where $I_{n-1}$ is the $n-1 times n-1$ identity. Then $||A|| = c$, so clearly $||A|| geq c$, but all its other eigenvalues are $varepsilon < c$.
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
$endgroup$
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
|
show 1 more comment
$begingroup$
No. Consider any diagonal matrix of the form $A = begin{pmatrix} c & 0 \ 0 & varepsilon I_{n-1} end{pmatrix}$ where $I_{n-1}$ is the $n-1 times n-1$ identity. Then $||A|| = c$, so clearly $||A|| geq c$, but all its other eigenvalues are $varepsilon < c$.
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
$endgroup$
No. Consider any diagonal matrix of the form $A = begin{pmatrix} c & 0 \ 0 & varepsilon I_{n-1} end{pmatrix}$ where $I_{n-1}$ is the $n-1 times n-1$ identity. Then $||A|| = c$, so clearly $||A|| geq c$, but all its other eigenvalues are $varepsilon < c$.
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
edited Jan 18 at 22:51
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
answered Jan 18 at 22:44
OldGodzillaOldGodzilla
58227
58227
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
|
show 1 more comment
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
1
1
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Could nilpotent matrix be invertible?
$endgroup$
– Jon
Jan 18 at 22:47
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Ah sorry, misread. I didn't realize that the matrix needed to be invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
Also, to answer your question, no. If a matrix has a zero eigenvalue then it is not invertible.
$endgroup$
– OldGodzilla
Jan 18 at 22:48
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
$begingroup$
@Jon Indeed it couldn't, for otherwise $A^n = 0$ would imply $I = (A^{-1})^n A^n = 0$.
$endgroup$
– MisterRiemann
Jan 18 at 22:49
1
1
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
$begingroup$
If you add $epsilon$ on the diagonal, it becomes invertible and constitutes a counter example for small enough $epsilon $
$endgroup$
– DLeMeur
Jan 18 at 22:51
|
show 1 more comment
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1
$begingroup$
What is $lVert ArVert$? Is it $lVert ArVert=sup_{lVert xrVert=1}lVert AxrVert?$
$endgroup$
– José Carlos Santos
Jan 18 at 22:43
$begingroup$
@JoséCarlosSantos Yes
$endgroup$
– Jon
Jan 18 at 22:44