Mixing
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
asked Jan 30 at 1:16
sejysejy
1589
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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
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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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
asked Jan 30 at 1:16
sejysejy
1589
$endgroup$
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.
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Related: math.stackexchange.com/questions/863063/….
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– StubbornAtom
Jan 30 at 16:39
add a comment |
$begingroup$
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
asked Jan 30 at 1:16
sejysejy
1589
$endgroup$
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
asked Jan 30 at 1:16
sejysejy
1589
asked Jan 30 at 1:16
sejysejy
1589
edited Jan 30 at 3:23
sejy
asked Jan 30 at 1:16
sejysejy
1589
asked Jan 30 at 1:16
sejysejy
1589
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.
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Related: math.stackexchange.com/questions/863063/….
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– StubbornAtom
Jan 30 at 16:39
add a comment |
$begingroup$
Related: math.stackexchange.com/questions/863063/….
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– StubbornAtom
Jan 30 at 16:39
$begingroup$
Related: math.stackexchange.com/questions/863063/….
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– StubbornAtom
Jan 30 at 16:39
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Related: math.stackexchange.com/questions/863063/….
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– StubbornAtom
Jan 30 at 16:39
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1 Answer
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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 $7$.
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
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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
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– Omnomnomnom
Jan 30 at 2:19
$begingroup$
Now I have written complete question
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– sejy
Jan 30 at 3:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $7$.
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $7$.
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $7$.
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
$endgroup$
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 $7$.
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
answered Jan 30 at 1:25
OmnomnomnomOmnomnomnom
129k793187
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Related: math.stackexchange.com/questions/863063/….
$endgroup$
– StubbornAtom
Jan 30 at 16:39