Let M be 10*10 matrix such that $M^{4}=0$ and $M^{i} neq 0$ for i=1,2,3 . Then what is the minimum and...












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Let $M$ be $10times10$ matrix such that $M^{4}=0$ and $M^{i} neq 0$ for $i=1,2,3$. Then what are the minimum and maximum ranks of the matrix. Where minimum rank and maximum rank is r and R respectively ?



a) R= 7, r=3



b)R= 6, r=4



c)R= 6, r=3



d)R= 7, r=4



think since there $M^{i}$ is not equal to $0$ for three values of $i$ so the minimum rank is $3$ But how to find the maximum rank?
I have not studied it's not part of curriculum that I have to solve without Jordan form.






linear-algebra matrices linear-transformations







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closed as off-topic by Alexander Gruber♦ Jan 30 at 1:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

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    Related: math.stackexchange.com/questions/863063/….
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    – StubbornAtom
    Jan 30 at 16:39
















0












$begingroup$


Let $M$ be $10times10$ matrix such that $M^{4}=0$ and $M^{i} neq 0$ for $i=1,2,3$. Then what are the minimum and maximum ranks of the matrix. Where minimum rank and maximum rank is r and R respectively ?



a) R= 7, r=3



b)R= 6, r=4



c)R= 6, r=3



d)R= 7, r=4



think since there $M^{i}$ is not equal to $0$ for three values of $i$ so the minimum rank is $3$ But how to find the maximum rank?
I have not studied it's not part of curriculum that I have to solve without Jordan form.






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$



closed as off-topic by Alexander Gruber♦ Jan 30 at 1:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Related: math.stackexchange.com/questions/863063/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 16:39














0












0








0


1



$begingroup$


Let $M$ be $10times10$ matrix such that $M^{4}=0$ and $M^{i} neq 0$ for $i=1,2,3$. Then what are the minimum and maximum ranks of the matrix. Where minimum rank and maximum rank is r and R respectively ?



a) R= 7, r=3



b)R= 6, r=4



c)R= 6, r=3



d)R= 7, r=4



think since there $M^{i}$ is not equal to $0$ for three values of $i$ so the minimum rank is $3$ But how to find the maximum rank?
I have not studied it's not part of curriculum that I have to solve without Jordan form.






linear-algebra matrices linear-transformations







share|cite|improve this question











$endgroup$




Let $M$ be $10times10$ matrix such that $M^{4}=0$ and $M^{i} neq 0$ for $i=1,2,3$. Then what are the minimum and maximum ranks of the matrix. Where minimum rank and maximum rank is r and R respectively ?



a) R= 7, r=3



b)R= 6, r=4



c)R= 6, r=3



d)R= 7, r=4



think since there $M^{i}$ is not equal to $0$ for three values of $i$ so the minimum rank is $3$ But how to find the maximum rank?
I have not studied it's not part of curriculum that I have to solve without Jordan form.






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 3:23







sejy

















asked Jan 30 at 1:16









sejysejy

1589




1589




closed as off-topic by Alexander Gruber♦ Jan 30 at 1:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Alexander Gruber♦ Jan 30 at 1:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Related: math.stackexchange.com/questions/863063/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 16:39


















  • $begingroup$
    Related: math.stackexchange.com/questions/863063/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 16:39
















$begingroup$
Related: math.stackexchange.com/questions/863063/….
$endgroup$
– StubbornAtom
Jan 30 at 16:39




$begingroup$
Related: math.stackexchange.com/questions/863063/….
$endgroup$
– StubbornAtom
Jan 30 at 16:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

Your answer for the minimal rank is correct.



I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $dim ker(A) = 10 - operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.



Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have not studied it's not part of curriculum that I have to solve without Jordan form .
    $endgroup$
    – sejy
    Jan 30 at 2:10








  • 2




    $begingroup$
    @sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
    $endgroup$
    – Omnomnomnom
    Jan 30 at 2:19










  • $begingroup$
    Now I have written complete question
    $endgroup$
    – sejy
    Jan 30 at 3:33




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Your answer for the minimal rank is correct.



I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $dim ker(A) = 10 - operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.



Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have not studied it's not part of curriculum that I have to solve without Jordan form .
    $endgroup$
    – sejy
    Jan 30 at 2:10








  • 2




    $begingroup$
    @sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
    $endgroup$
    – Omnomnomnom
    Jan 30 at 2:19










  • $begingroup$
    Now I have written complete question
    $endgroup$
    – sejy
    Jan 30 at 3:33


















2












$begingroup$

Your answer for the minimal rank is correct.



I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $dim ker(A) = 10 - operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.



Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have not studied it's not part of curriculum that I have to solve without Jordan form .
    $endgroup$
    – sejy
    Jan 30 at 2:10








  • 2




    $begingroup$
    @sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
    $endgroup$
    – Omnomnomnom
    Jan 30 at 2:19










  • $begingroup$
    Now I have written complete question
    $endgroup$
    – sejy
    Jan 30 at 3:33
















2












2








2





$begingroup$

Your answer for the minimal rank is correct.



I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $dim ker(A) = 10 - operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.



Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.






share|cite|improve this answer









$endgroup$



Your answer for the minimal rank is correct.



I find it easiest to think of this question in terms of the Jordan form of $M$. Because Because $M$ has minimal polynomial $p(x) = x^4$, we know that $M$ has $0$ as its only eigenvalue, and that the largest Jordan block has size $4$. On the other hand, $dim ker(A) = 10 - operatorname{rank}(A)$ is the total number of Jordan blocks associated with zero.



Thus, our question amounts to finding the Jordan form matrix (with only zero-blocks) that contains the fewest blocks, each block of size at most $4$. Since $M$ must contain at least $10/4 = 2.5$ such blocks, we conclude that the kernel has dimension at least $3$, which is to say that the rank of $M$ is at most $7$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 30 at 1:25









OmnomnomnomOmnomnomnom

129k793187




129k793187












  • $begingroup$
    I have not studied it's not part of curriculum that I have to solve without Jordan form .
    $endgroup$
    – sejy
    Jan 30 at 2:10








  • 2




    $begingroup$
    @sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
    $endgroup$
    – Omnomnomnom
    Jan 30 at 2:19










  • $begingroup$
    Now I have written complete question
    $endgroup$
    – sejy
    Jan 30 at 3:33




















  • $begingroup$
    I have not studied it's not part of curriculum that I have to solve without Jordan form .
    $endgroup$
    – sejy
    Jan 30 at 2:10








  • 2




    $begingroup$
    @sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
    $endgroup$
    – Omnomnomnom
    Jan 30 at 2:19










  • $begingroup$
    Now I have written complete question
    $endgroup$
    – sejy
    Jan 30 at 3:33


















$begingroup$
I have not studied it's not part of curriculum that I have to solve without Jordan form .
$endgroup$
– sejy
Jan 30 at 2:10






$begingroup$
I have not studied it's not part of curriculum that I have to solve without Jordan form .
$endgroup$
– sejy
Jan 30 at 2:10






2




2




$begingroup$
@sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
$endgroup$
– Omnomnomnom
Jan 30 at 2:19




$begingroup$
@sejy I think you could also make a proof based on Sylvester's rank inequality, that is $$ operatorname{rank}(AB) geq operatorname{rank}(A) + operatorname{rank}(B) - n $$ it would be easier to give you a helpful answer, though, if you edited your question to include more context
$endgroup$
– Omnomnomnom
Jan 30 at 2:19












$begingroup$
Now I have written complete question
$endgroup$
– sejy
Jan 30 at 3:33






$begingroup$
Now I have written complete question
$endgroup$
– sejy
Jan 30 at 3:33





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