The dimension of $H_{n} $ ,the space of all $ n times n $ matrices, as a vector space over $ mathbb{R} $ , is
$begingroup$
Let $ n $ be a positive integer and let $ H_{n} $ be the space of all $ n times n $ matrices
$ A = (a_{ij}) $ with entries in $ mathbb{R} $ satisfying $ a_{ij} = a_{rs} $ whenever $ i + j = r + s ,, (i,j,r,s=1,2,dots,n)$ . Then the dimension of $ H_{n} $ , as a vector space over $ mathbb{R} $ , is
$ 1. quad n^{2} hspace{50pt} 2.quad n^{2}-n+1 hspace{50pt} 3. quad 2n+1 hspace{50pt} 4.quad 2n-1 $
I am trying to solve it by taking $ n=3 $ for generalized form of matrix and found
M=
$ a_{11}$ $a_{12}$ $a_{13}$
$ a_{21}$ $ a_{22} $ $ a_{23}$
$ a_{31}$ $a_{32} $ $a_{33}$
where $ a_{12} = a_{21} $ , $ a_{13} = a_{31}= a_{22} $ , $ a_{23} = a_{32} $
Now I can't understand how to conclude it?
Can I conclude it as follows:
Since the matrix has $ 5 $ non-trivial distinct elements, so the dimension of $ H_{n} $ is $ 5 $ and as $ 5= 2 * 3 -1 $ so ant answer is $ 2n -1 $ ?
linear-algebra
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
$endgroup$
|
show 3 more comments
$begingroup$
Let $ n $ be a positive integer and let $ H_{n} $ be the space of all $ n times n $ matrices
$ A = (a_{ij}) $ with entries in $ mathbb{R} $ satisfying $ a_{ij} = a_{rs} $ whenever $ i + j = r + s ,, (i,j,r,s=1,2,dots,n)$ . Then the dimension of $ H_{n} $ , as a vector space over $ mathbb{R} $ , is
$ 1. quad n^{2} hspace{50pt} 2.quad n^{2}-n+1 hspace{50pt} 3. quad 2n+1 hspace{50pt} 4.quad 2n-1 $
I am trying to solve it by taking $ n=3 $ for generalized form of matrix and found
M=
$ a_{11}$ $a_{12}$ $a_{13}$
$ a_{21}$ $ a_{22} $ $ a_{23}$
$ a_{31}$ $a_{32} $ $a_{33}$
where $ a_{12} = a_{21} $ , $ a_{13} = a_{31}= a_{22} $ , $ a_{23} = a_{32} $
Now I can't understand how to conclude it?
Can I conclude it as follows:
Since the matrix has $ 5 $ non-trivial distinct elements, so the dimension of $ H_{n} $ is $ 5 $ and as $ 5= 2 * 3 -1 $ so ant answer is $ 2n -1 $ ?
linear-algebra
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
$endgroup$
$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
1
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
1
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57
|
show 3 more comments
$begingroup$
Let $ n $ be a positive integer and let $ H_{n} $ be the space of all $ n times n $ matrices
$ A = (a_{ij}) $ with entries in $ mathbb{R} $ satisfying $ a_{ij} = a_{rs} $ whenever $ i + j = r + s ,, (i,j,r,s=1,2,dots,n)$ . Then the dimension of $ H_{n} $ , as a vector space over $ mathbb{R} $ , is
$ 1. quad n^{2} hspace{50pt} 2.quad n^{2}-n+1 hspace{50pt} 3. quad 2n+1 hspace{50pt} 4.quad 2n-1 $
I am trying to solve it by taking $ n=3 $ for generalized form of matrix and found
M=
$ a_{11}$ $a_{12}$ $a_{13}$
$ a_{21}$ $ a_{22} $ $ a_{23}$
$ a_{31}$ $a_{32} $ $a_{33}$
where $ a_{12} = a_{21} $ , $ a_{13} = a_{31}= a_{22} $ , $ a_{23} = a_{32} $
Now I can't understand how to conclude it?
Can I conclude it as follows:
Since the matrix has $ 5 $ non-trivial distinct elements, so the dimension of $ H_{n} $ is $ 5 $ and as $ 5= 2 * 3 -1 $ so ant answer is $ 2n -1 $ ?
linear-algebra
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
$endgroup$
Let $ n $ be a positive integer and let $ H_{n} $ be the space of all $ n times n $ matrices
$ A = (a_{ij}) $ with entries in $ mathbb{R} $ satisfying $ a_{ij} = a_{rs} $ whenever $ i + j = r + s ,, (i,j,r,s=1,2,dots,n)$ . Then the dimension of $ H_{n} $ , as a vector space over $ mathbb{R} $ , is
$ 1. quad n^{2} hspace{50pt} 2.quad n^{2}-n+1 hspace{50pt} 3. quad 2n+1 hspace{50pt} 4.quad 2n-1 $
I am trying to solve it by taking $ n=3 $ for generalized form of matrix and found
M=
$ a_{11}$ $a_{12}$ $a_{13}$
$ a_{21}$ $ a_{22} $ $ a_{23}$
$ a_{31}$ $a_{32} $ $a_{33}$
where $ a_{12} = a_{21} $ , $ a_{13} = a_{31}= a_{22} $ , $ a_{23} = a_{32} $
Now I can't understand how to conclude it?
Can I conclude it as follows:
Since the matrix has $ 5 $ non-trivial distinct elements, so the dimension of $ H_{n} $ is $ 5 $ and as $ 5= 2 * 3 -1 $ so ant answer is $ 2n -1 $ ?
linear-algebra
linear-algebra
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
edited Dec 7 '18 at 16:00
Scientifica
6,62141335
6,62141335
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
asked Dec 7 '18 at 15:13
N. MasantaN. Masanta
94
94
$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
1
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
1
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57
|
show 3 more comments
$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
1
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
1
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57
$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
1
1
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
1
1
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider the diagonals of the matrix going form lower left to upper right. All the entries in those diagonals are of the form $a_{ij}$, where i+j=k for some fixed k. Count the number of such diagonals and those are the equivalence classes of entries.
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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$begingroup$
Consider the diagonals of the matrix going form lower left to upper right. All the entries in those diagonals are of the form $a_{ij}$, where i+j=k for some fixed k. Count the number of such diagonals and those are the equivalence classes of entries.
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
add a comment |
$begingroup$
Consider the diagonals of the matrix going form lower left to upper right. All the entries in those diagonals are of the form $a_{ij}$, where i+j=k for some fixed k. Count the number of such diagonals and those are the equivalence classes of entries.
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
$endgroup$
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
add a comment |
$begingroup$
Consider the diagonals of the matrix going form lower left to upper right. All the entries in those diagonals are of the form $a_{ij}$, where i+j=k for some fixed k. Count the number of such diagonals and those are the equivalence classes of entries.
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
$endgroup$
Consider the diagonals of the matrix going form lower left to upper right. All the entries in those diagonals are of the form $a_{ij}$, where i+j=k for some fixed k. Count the number of such diagonals and those are the equivalence classes of entries.
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
answered Dec 7 '18 at 15:49
Joel PereiraJoel Pereira
74519
74519
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
add a comment |
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
Is it the general rules for finding dimension of the such kind of vector space? If it happens, then please mention the proper reference. Thank you.
$endgroup$
– N. Masanta
Dec 7 '18 at 15:54
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
$begingroup$
The general rule is to find the number of degrees of freedom you have. In this case, once you found a$_{ij}$, the entries on the same upper diagonal were determined.
$endgroup$
– Joel Pereira
Dec 7 '18 at 19:11
add a comment |
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$begingroup$
I'd count it for $n=2,3$.
$endgroup$
– Berci
Dec 7 '18 at 15:15
1
$begingroup$
How many equivalence classes $(i,j)$ are there?
$endgroup$
– Federico
Dec 7 '18 at 15:15
1
$begingroup$
Why don't you try with some small examples? Look at $n = 2$, write some matrices down and see if you can guess a basis.
$endgroup$
– ryan221b
Dec 7 '18 at 15:15
$begingroup$
Regarding your edit, making a conclusion from a single exemple is really dangerous, I almost concluded that the answer was $n^2-n+1$ because it was correct for $n=1,2$.
$endgroup$
– F.Carette
Dec 7 '18 at 15:57
$begingroup$
I am trying to solve it by taking n = 3 for generalized form and found M = a11 a12 a13 a21 a22 a23 a31 a32 a33 where a12 = a21 , a13 = a31 = a22 , a23 = a32 Now I can't understand how to conclude it? Can I conclude it as follows: Since the matrix has 5 non-trivial distinct elements, so the dimension of H_n is 5 and as 5 = 2 * 3 - 1 so ant answer is 2n - 1 ?
$endgroup$
– N. Masanta
Dec 7 '18 at 15:57