How can we show that $(I-A)$ is invertible?
$begingroup$
$A$ is an $ntimes n$ matrix with $|A|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $|(I-A)^{-1}|lefrac1{(1-a)}$.
It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
matrices
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
$endgroup$
add a comment |
$begingroup$
$A$ is an $ntimes n$ matrix with $|A|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $|(I-A)^{-1}|lefrac1{(1-a)}$.
It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
matrices
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
$endgroup$
1
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50
add a comment |
$begingroup$
$A$ is an $ntimes n$ matrix with $|A|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $|(I-A)^{-1}|lefrac1{(1-a)}$.
It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
matrices
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
$endgroup$
$A$ is an $ntimes n$ matrix with $|A|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $|(I-A)^{-1}|lefrac1{(1-a)}$.
It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?
matrices
matrices
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
edited Feb 17 '12 at 11:14
Asaf Karagila♦
303k32429761
303k32429761
asked Feb 17 '12 at 6:33
roserose
9315
asked Feb 17 '12 at 6:33
roserose
9315
asked Feb 17 '12 at 6:33
roserose
9315
9315
1
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50
add a comment |
1
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50
1
1
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint: Show that a certain series converges in the norm $|cdot |$ and that this is an inverse for $I-A$.
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $|x| = 1$, so $|A| ge 1$.
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
$bf Hint$: Consider the series $sum_{n=0}^infty A^n$
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
$endgroup$
add a comment |
$begingroup$
Another way to show this is the following:
$I-A$ is invertible iff $ker(I-A) = {0}$.
Now, let $x in mathbb{R}^nbackslash { 0 }$. Then
$$
lVert x rVert leq lVert x-AxrVert + lVert Ax rVert leq lVert (I - A) x rVert +lVert A rVert lVert x rVert underset{text{Assumption}}{<} lVert (I - A) x rVert + lVert x rVert
$$
So $0 < lVert (I-A) x rVert$ and thus $x notin ker(I-A)$.
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Show that a certain series converges in the norm $|cdot |$ and that this is an inverse for $I-A$.
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
Hint: Show that a certain series converges in the norm $|cdot |$ and that this is an inverse for $I-A$.
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
Hint: Show that a certain series converges in the norm $|cdot |$ and that this is an inverse for $I-A$.
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
$endgroup$
Hint: Show that a certain series converges in the norm $|cdot |$ and that this is an inverse for $I-A$.
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
answered Feb 17 '12 at 6:40
Robert IsraelRobert Israel
321k23210462
321k23210462
add a comment |
add a comment |
$begingroup$
You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $|x| = 1$, so $|A| ge 1$.
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $|x| = 1$, so $|A| ge 1$.
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $|x| = 1$, so $|A| ge 1$.
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
$endgroup$
You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $lambda = 1$ is not an eigenvalue of $A$.
For a contradiction, assume $lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $|x| = 1$, so $|A| ge 1$.
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
answered Feb 17 '12 at 11:12
mrfmrf
37.4k54685
37.4k54685
add a comment |
add a comment |
$begingroup$
$bf Hint$: Consider the series $sum_{n=0}^infty A^n$
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
$endgroup$
add a comment |
$begingroup$
$bf Hint$: Consider the series $sum_{n=0}^infty A^n$
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
$endgroup$
add a comment |
$begingroup$
$bf Hint$: Consider the series $sum_{n=0}^infty A^n$
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
$endgroup$
$bf Hint$: Consider the series $sum_{n=0}^infty A^n$
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
answered Feb 17 '12 at 6:39
azarelazarel
11.1k22431
11.1k22431
add a comment |
add a comment |
$begingroup$
Another way to show this is the following:
$I-A$ is invertible iff $ker(I-A) = {0}$.
Now, let $x in mathbb{R}^nbackslash { 0 }$. Then
$$
lVert x rVert leq lVert x-AxrVert + lVert Ax rVert leq lVert (I - A) x rVert +lVert A rVert lVert x rVert underset{text{Assumption}}{<} lVert (I - A) x rVert + lVert x rVert
$$
So $0 < lVert (I-A) x rVert$ and thus $x notin ker(I-A)$.
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
$begingroup$
Another way to show this is the following:
$I-A$ is invertible iff $ker(I-A) = {0}$.
Now, let $x in mathbb{R}^nbackslash { 0 }$. Then
$$
lVert x rVert leq lVert x-AxrVert + lVert Ax rVert leq lVert (I - A) x rVert +lVert A rVert lVert x rVert underset{text{Assumption}}{<} lVert (I - A) x rVert + lVert x rVert
$$
So $0 < lVert (I-A) x rVert$ and thus $x notin ker(I-A)$.
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
$endgroup$
add a comment |
$begingroup$
Another way to show this is the following:
$I-A$ is invertible iff $ker(I-A) = {0}$.
Now, let $x in mathbb{R}^nbackslash { 0 }$. Then
$$
lVert x rVert leq lVert x-AxrVert + lVert Ax rVert leq lVert (I - A) x rVert +lVert A rVert lVert x rVert underset{text{Assumption}}{<} lVert (I - A) x rVert + lVert x rVert
$$
So $0 < lVert (I-A) x rVert$ and thus $x notin ker(I-A)$.
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
$endgroup$
Another way to show this is the following:
$I-A$ is invertible iff $ker(I-A) = {0}$.
Now, let $x in mathbb{R}^nbackslash { 0 }$. Then
$$
lVert x rVert leq lVert x-AxrVert + lVert Ax rVert leq lVert (I - A) x rVert +lVert A rVert lVert x rVert underset{text{Assumption}}{<} lVert (I - A) x rVert + lVert x rVert
$$
So $0 < lVert (I-A) x rVert$ and thus $x notin ker(I-A)$.
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
answered Nov 27 '18 at 15:58
bruderjakob17bruderjakob17
1997
1997
add a comment |
add a comment |
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1
$begingroup$
Do you know of the geometric series?
$endgroup$
– Mariano Suárez-Álvarez
Feb 17 '12 at 6:39
$begingroup$
yes I know geometric series
$endgroup$
– rose
Feb 17 '12 at 6:50