Mixing
Maximum number of zeros in a diagonal matrix of order n [closed]
$begingroup$
According to my book, minimum number of zeros is $n(n-1)$
And maximum number of zeros is $n^2-1$.
I don't get the maximum one.
Edit: guys I'm sorry I didn't read the part in my book which states a diagonal matrix has at least one element in the diagonal as non zero
matrices
asked Jan 2 at 14:47
S. SuryaS. Surya
143
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closed as off-topic by A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser Jan 3 at 4:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
According to my book, minimum number of zeros is $n(n-1)$
And maximum number of zeros is $n^2-1$.
I don't get the maximum one.
Edit: guys I'm sorry I didn't read the part in my book which states a diagonal matrix has at least one element in the diagonal as non zero
matrices
asked Jan 2 at 14:47
S. SuryaS. Surya
143
$endgroup$
closed as off-topic by A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser Jan 3 at 4:40
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." – A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
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– Ethan Bolker
Jan 2 at 14:49
add a comment |
$begingroup$
According to my book, minimum number of zeros is $n(n-1)$
And maximum number of zeros is $n^2-1$.
I don't get the maximum one.
Edit: guys I'm sorry I didn't read the part in my book which states a diagonal matrix has at least one element in the diagonal as non zero
matrices
asked Jan 2 at 14:47
S. SuryaS. Surya
143
$endgroup$
According to my book, minimum number of zeros is $n(n-1)$
And maximum number of zeros is $n^2-1$.
I don't get the maximum one.
Edit: guys I'm sorry I didn't read the part in my book which states a diagonal matrix has at least one element in the diagonal as non zero
matrices
matrices
asked Jan 2 at 14:47
S. SuryaS. Surya
143
asked Jan 2 at 14:47
S. SuryaS. Surya
143
edited Jan 2 at 15:10
S. Surya
asked Jan 2 at 14:47
S. SuryaS. Surya
143
asked Jan 2 at 14:47
S. SuryaS. Surya
143
asked Jan 2 at 14:47
S. SuryaS. Surya
143
143
closed as off-topic by A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser Jan 3 at 4:40
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." – A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser Jan 3 at 4:40
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." – A. Pongrácz, José Carlos Santos, Crostul, Lord Shark the Unknown, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
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Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
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– Ethan Bolker
Jan 2 at 14:49
add a comment |
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Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
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– Ethan Bolker
Jan 2 at 14:49
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Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
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– Ethan Bolker
Jan 2 at 14:49
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Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
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– Ethan Bolker
Jan 2 at 14:49
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1 Answer
1
active
oldest
votes
$begingroup$
According to this definition the numbers can be zero in the diagonal, then the maximum number of zeros in a diagonal matrix is $n^2$. But according of the definition of your book the maximum number of zeros in a diagonal matrix is $n^2-1$. In my opinion, this is only subtleties of the definition without much relevance.
answered Jan 2 at 14:54
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to this definition the numbers can be zero in the diagonal, then the maximum number of zeros in a diagonal matrix is $n^2$. But according of the definition of your book the maximum number of zeros in a diagonal matrix is $n^2-1$. In my opinion, this is only subtleties of the definition without much relevance.
answered Jan 2 at 14:54
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
add a comment |
$begingroup$
According to this definition the numbers can be zero in the diagonal, then the maximum number of zeros in a diagonal matrix is $n^2$. But according of the definition of your book the maximum number of zeros in a diagonal matrix is $n^2-1$. In my opinion, this is only subtleties of the definition without much relevance.
answered Jan 2 at 14:54
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
add a comment |
$begingroup$
According to this definition the numbers can be zero in the diagonal, then the maximum number of zeros in a diagonal matrix is $n^2$. But according of the definition of your book the maximum number of zeros in a diagonal matrix is $n^2-1$. In my opinion, this is only subtleties of the definition without much relevance.
answered Jan 2 at 14:54
El PastaEl Pasta
34515
$endgroup$
According to this definition the numbers can be zero in the diagonal, then the maximum number of zeros in a diagonal matrix is $n^2$. But according of the definition of your book the maximum number of zeros in a diagonal matrix is $n^2-1$. In my opinion, this is only subtleties of the definition without much relevance.
answered Jan 2 at 14:54
El PastaEl Pasta
34515
edited Jan 2 at 15:12
answered Jan 2 at 14:54
El PastaEl Pasta
34515
answered Jan 2 at 14:54
El PastaEl Pasta
34515
answered Jan 2 at 14:54
El PastaEl Pasta
34515
34515
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
add a comment |
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
1
1
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
I'm guessing that OP's definition may require at least one nonzero entry on the main diagonal, hence his $n^2-1$ rather than $n^2$. But I would agree with your assessment, +1.
$endgroup$
– MPW
Jan 2 at 15:03
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
We agree, it's a question of definitions according to my opinion.
$endgroup$
– El Pasta
Jan 2 at 15:06
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
If I had to take a long shot, I'd guess the problem said "In a nonzero order-$n$ diagonal matrix, what are the max and min number of zeroes?"
$endgroup$
– John Hughes
Jan 2 at 15:08
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
$begingroup$
Depends the definition I think, but I think the min is always $n^2-n=n(n-1)$.
$endgroup$
– El Pasta
Jan 2 at 15:34
add a comment |
$begingroup$
Welcome to stackexchange. Something is missing from the statement of your problem. As you've asked it, you could have all $0$'s or none. Please edit the question (don't explain in a comment). Use mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 2 at 14:49