Let $A$ be $10 times 10$ real matrix. then which of the following is correct?
$begingroup$
Let $A$ be $10 times 10$ real matrix. then which of the following is correct?
[$rho(A)=Rank(A)]$
(A) $rho(A^8)=rho(A^9)$
(B)$rho(A^9)=rho(A^{10})$
(C) $rho(A^{10})=rho(A^{11})$
(D)$rho(A^8)=rho(A^7)$
(A),(B),(D) are false. since if I take an upper triangular matrix $A$ with all zero entries in the diagonal and 1 in all $i<j$ region. Then, (A),(B),(D) are false. Hence, $C$ is the correct answer. Is there any shorter way to solve this?
real-analysis sequences-and-series limits alternative-proof
edited Dec 9 '18 at 6:00
Shaun
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
$endgroup$
add a comment |
$begingroup$
Let $A$ be $10 times 10$ real matrix. then which of the following is correct?
[$rho(A)=Rank(A)]$
(A) $rho(A^8)=rho(A^9)$
(B)$rho(A^9)=rho(A^{10})$
(C) $rho(A^{10})=rho(A^{11})$
(D)$rho(A^8)=rho(A^7)$
(A),(B),(D) are false. since if I take an upper triangular matrix $A$ with all zero entries in the diagonal and 1 in all $i<j$ region. Then, (A),(B),(D) are false. Hence, $C$ is the correct answer. Is there any shorter way to solve this?
real-analysis sequences-and-series limits alternative-proof
edited Dec 9 '18 at 6:00
Shaun
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
$endgroup$
$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23
add a comment |
$begingroup$
Let $A$ be $10 times 10$ real matrix. then which of the following is correct?
[$rho(A)=Rank(A)]$
(A) $rho(A^8)=rho(A^9)$
(B)$rho(A^9)=rho(A^{10})$
(C) $rho(A^{10})=rho(A^{11})$
(D)$rho(A^8)=rho(A^7)$
(A),(B),(D) are false. since if I take an upper triangular matrix $A$ with all zero entries in the diagonal and 1 in all $i<j$ region. Then, (A),(B),(D) are false. Hence, $C$ is the correct answer. Is there any shorter way to solve this?
real-analysis sequences-and-series limits alternative-proof
edited Dec 9 '18 at 6:00
Shaun
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
$endgroup$
Let $A$ be $10 times 10$ real matrix. then which of the following is correct?
[$rho(A)=Rank(A)]$
(A) $rho(A^8)=rho(A^9)$
(B)$rho(A^9)=rho(A^{10})$
(C) $rho(A^{10})=rho(A^{11})$
(D)$rho(A^8)=rho(A^7)$
(A),(B),(D) are false. since if I take an upper triangular matrix $A$ with all zero entries in the diagonal and 1 in all $i<j$ region. Then, (A),(B),(D) are false. Hence, $C$ is the correct answer. Is there any shorter way to solve this?
real-analysis sequences-and-series limits alternative-proof
real-analysis sequences-and-series limits alternative-proof
edited Dec 9 '18 at 6:00
Shaun
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
edited Dec 9 '18 at 6:00
Shaun
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
edited Dec 9 '18 at 6:00
Shaun
9,010113682
edited Dec 9 '18 at 6:00
Shaun
9,010113682
edited Dec 9 '18 at 6:00
Shaun
9,010113682
9,010113682
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
asked Dec 9 '18 at 5:44
Unknown xUnknown x
2,50511026
2,50511026
$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23
add a comment |
$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23
$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If there is $v$ such that $A^{11} v = 0$ but $A^{10} v ne 0$, then the null spaces of $A, A^2, ldots, A^{11}$ are all distinct. But these are nested, so their dimensions are all different. $mathfrak N(A)$ has dimension at least $1$, ...., $mathfrak N(A^{11})$ has dimension at least $11$, but that's impossible.
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
$endgroup$
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
add a comment |
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1 Answer
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$begingroup$
If there is $v$ such that $A^{11} v = 0$ but $A^{10} v ne 0$, then the null spaces of $A, A^2, ldots, A^{11}$ are all distinct. But these are nested, so their dimensions are all different. $mathfrak N(A)$ has dimension at least $1$, ...., $mathfrak N(A^{11})$ has dimension at least $11$, but that's impossible.
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
$endgroup$
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
add a comment |
$begingroup$
If there is $v$ such that $A^{11} v = 0$ but $A^{10} v ne 0$, then the null spaces of $A, A^2, ldots, A^{11}$ are all distinct. But these are nested, so their dimensions are all different. $mathfrak N(A)$ has dimension at least $1$, ...., $mathfrak N(A^{11})$ has dimension at least $11$, but that's impossible.
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
$endgroup$
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
add a comment |
$begingroup$
If there is $v$ such that $A^{11} v = 0$ but $A^{10} v ne 0$, then the null spaces of $A, A^2, ldots, A^{11}$ are all distinct. But these are nested, so their dimensions are all different. $mathfrak N(A)$ has dimension at least $1$, ...., $mathfrak N(A^{11})$ has dimension at least $11$, but that's impossible.
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
$endgroup$
If there is $v$ such that $A^{11} v = 0$ but $A^{10} v ne 0$, then the null spaces of $A, A^2, ldots, A^{11}$ are all distinct. But these are nested, so their dimensions are all different. $mathfrak N(A)$ has dimension at least $1$, ...., $mathfrak N(A^{11})$ has dimension at least $11$, but that's impossible.
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
answered Dec 9 '18 at 6:05
Robert IsraelRobert Israel
321k23210463
321k23210463
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
add a comment |
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
I know null spaces are nested. But I am not able to prove How these null spaces are distinct? is it followed from the fact that $A^{10}(Av)=0$ but $A^9(Av)neq 0$. So, $mathfrak N(A^{10})neqmathfrak N(A^{9})$. so, on.am I correct?
$endgroup$
– Unknown x
Dec 9 '18 at 6:24
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
$begingroup$
Yes, that's right.
$endgroup$
– Robert Israel
Dec 9 '18 at 17:09
add a comment |
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$begingroup$
Not really. The Jordan form is useful here.
$endgroup$
– copper.hat
Dec 9 '18 at 6:10
$begingroup$
How Jordan form is useful here?
$endgroup$
– Unknown x
Dec 9 '18 at 6:13
$begingroup$
The rank can be seen easily from the Jordan form (the Jordan blocks corresponding to the zero eigenvalues are the only places where the rank changes when taking powers). So, it is straightforward, by taking Jordan blocks of the appropriate size, to see that A,B,D fail. It is also straightforward to see that the rank cannot change after the $n$th power (because all Jordan blocks corresponding to zero eigenvalues will be zero at this stage).
$endgroup$
– copper.hat
Dec 9 '18 at 6:23