Let $A$ be $10 times 10$ real matrix. then which of the following is correct?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 6:00









Shaun

9,088113683




9,088113683










asked Dec 9 '18 at 5:44









Unknown xUnknown x

2,51911026




2,51911026












  • $begingroup$
    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










  • $begingroup$
    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












1 Answer
1






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oldest

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3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 6:05









Robert IsraelRobert Israel

323k23212466




323k23212466












  • $begingroup$
    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












  • $begingroup$
    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


















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