Mixing
Dimensions of symmetric and skew-symmetric matrices
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Let $textbf A$ denote the space of symmetric $(ntimes n)$ matrices over the field $mathbb K$, and $textbf B$ the space of skew-symmetric $(ntimes n)$ matrices over the field $mathbb K$. Then $dim (textbf A)=n(n+1)/2$ and $dim (textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $dim$ refers to linear spaces.
linear-algebra matrices
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
$endgroup$
add a comment |
$begingroup$
Let $textbf A$ denote the space of symmetric $(ntimes n)$ matrices over the field $mathbb K$, and $textbf B$ the space of skew-symmetric $(ntimes n)$ matrices over the field $mathbb K$. Then $dim (textbf A)=n(n+1)/2$ and $dim (textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $dim$ refers to linear spaces.
linear-algebra matrices
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
$endgroup$
1
$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
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I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
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You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
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And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40
add a comment |
$begingroup$
Let $textbf A$ denote the space of symmetric $(ntimes n)$ matrices over the field $mathbb K$, and $textbf B$ the space of skew-symmetric $(ntimes n)$ matrices over the field $mathbb K$. Then $dim (textbf A)=n(n+1)/2$ and $dim (textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $dim$ refers to linear spaces.
linear-algebra matrices
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
$endgroup$
Let $textbf A$ denote the space of symmetric $(ntimes n)$ matrices over the field $mathbb K$, and $textbf B$ the space of skew-symmetric $(ntimes n)$ matrices over the field $mathbb K$. Then $dim (textbf A)=n(n+1)/2$ and $dim (textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $dim$ refers to linear spaces.
linear-algebra matrices
linear-algebra matrices
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
edited Aug 23 '12 at 12:38
Marc van Leeuwen
88.3k5111228
88.3k5111228
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
asked Aug 23 '12 at 9:55
Christian IvicevicChristian Ivicevic
1,73332234
1,73332234
1
$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
$begingroup$
I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
$begingroup$
You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
$begingroup$
And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40
add a comment |
1
$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
$begingroup$
I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
$begingroup$
You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
$begingroup$
And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40
1
1
$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
$begingroup$
I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
$begingroup$
I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
$begingroup$
You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
$begingroup$
You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
$begingroup$
And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40
$begingroup$
And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
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add a comment |
$begingroup$
Here is my two cents:
begin{eqnarray}
M_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
*&*&*&*&cdots \
*&*&*&*& \
*&*&*&*& \
*&*&*&*& \
vdots&&&&ddots
end{pmatrix} hspace{.5cm} text{with $n^2$ elements}\ \ \
Skew_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix}
end{eqnarray}
For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $dim(Skew_{n times n}(mathbb{R}) + Sym_{n times n}(mathbb{R})) = dim(M_{n times n}(mathbb{R}))$ and $dim(Skew_{n times n}(mathbb{R}))=frac{n^2-n}{2}$ then we have that
begin{eqnarray}
frac{n^2-n}{2}+dim(Sym_{n times n}(mathbb{R})))=n^2
end{eqnarray}
or
begin{eqnarray}
dim(Sym_{n times n}(mathbb{R})))=frac{n^2+n}{2}.
end{eqnarray}
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
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add a comment |
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This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n \ 2 end{array} right) = frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct symbols, where now a symbol is either an index ($1 , ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n+1 \ 2 end{array} right) = frac{n(n+1)}{2}$ such sets.
answered Jul 9 '14 at 19:37
StirlingStirling
36429
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add a comment |
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The dimension of symmetric matrices is $frac{n(n+1)}2$ because they have one basis as the matrices ${M_{ij}}_{n ge i ge j ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is ${M_{ij}}_{n ge i > j ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
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But this is no explanation why the symmetric matrices have the specified $dim$.
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– Christian Ivicevic
Aug 23 '12 at 10:03
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Do you mean $n^2$?
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– enzotib
Aug 23 '12 at 10:04
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Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
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– Rijul Saini
Aug 23 '12 at 10:07
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@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
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– Christian Ivicevic
Aug 23 '12 at 10:13
add a comment |
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There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$begin{matrix}
begin{pmatrix}
0 &*' \
*& 0 \
end{pmatrix} & *
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' \
*& 0 & *' \
* & * & 0 \
end{pmatrix} & begin{matrix}
*& \
* & *
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' &*'\
*& 0 & *' &*'\
* & * & 0 &*'\
* & * &* & 0
end{pmatrix} & begin{matrix}
* & & \
*&* & \
* & * &*
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix} & begin{matrix}
*& \
*&*& \
*&*&*& \
vdots&&&&ddots
end{matrix}
end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + dots + (n-1).$$
So by Gauss, we have $$frac{(n-1)(n-1+1)}{2} = frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$frac{(n-1)n}{2} + n = frac{n^2 + n}{2}$$
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
$endgroup$
add a comment |
$begingroup$
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
$endgroup$
add a comment |
$begingroup$
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
$endgroup$
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
edited May 9 '15 at 15:20
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
answered Aug 23 '12 at 10:10
enzotibenzotib
5,84821531
5,84821531
add a comment |
add a comment |
$begingroup$
Here is my two cents:
begin{eqnarray}
M_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
*&*&*&*&cdots \
*&*&*&*& \
*&*&*&*& \
*&*&*&*& \
vdots&&&&ddots
end{pmatrix} hspace{.5cm} text{with $n^2$ elements}\ \ \
Skew_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix}
end{eqnarray}
For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $dim(Skew_{n times n}(mathbb{R}) + Sym_{n times n}(mathbb{R})) = dim(M_{n times n}(mathbb{R}))$ and $dim(Skew_{n times n}(mathbb{R}))=frac{n^2-n}{2}$ then we have that
begin{eqnarray}
frac{n^2-n}{2}+dim(Sym_{n times n}(mathbb{R})))=n^2
end{eqnarray}
or
begin{eqnarray}
dim(Sym_{n times n}(mathbb{R})))=frac{n^2+n}{2}.
end{eqnarray}
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
$endgroup$
add a comment |
$begingroup$
Here is my two cents:
begin{eqnarray}
M_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
*&*&*&*&cdots \
*&*&*&*& \
*&*&*&*& \
*&*&*&*& \
vdots&&&&ddots
end{pmatrix} hspace{.5cm} text{with $n^2$ elements}\ \ \
Skew_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix}
end{eqnarray}
For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $dim(Skew_{n times n}(mathbb{R}) + Sym_{n times n}(mathbb{R})) = dim(M_{n times n}(mathbb{R}))$ and $dim(Skew_{n times n}(mathbb{R}))=frac{n^2-n}{2}$ then we have that
begin{eqnarray}
frac{n^2-n}{2}+dim(Sym_{n times n}(mathbb{R})))=n^2
end{eqnarray}
or
begin{eqnarray}
dim(Sym_{n times n}(mathbb{R})))=frac{n^2+n}{2}.
end{eqnarray}
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
$endgroup$
add a comment |
$begingroup$
Here is my two cents:
begin{eqnarray}
M_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
*&*&*&*&cdots \
*&*&*&*& \
*&*&*&*& \
*&*&*&*& \
vdots&&&&ddots
end{pmatrix} hspace{.5cm} text{with $n^2$ elements}\ \ \
Skew_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix}
end{eqnarray}
For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $dim(Skew_{n times n}(mathbb{R}) + Sym_{n times n}(mathbb{R})) = dim(M_{n times n}(mathbb{R}))$ and $dim(Skew_{n times n}(mathbb{R}))=frac{n^2-n}{2}$ then we have that
begin{eqnarray}
frac{n^2-n}{2}+dim(Sym_{n times n}(mathbb{R})))=n^2
end{eqnarray}
or
begin{eqnarray}
dim(Sym_{n times n}(mathbb{R})))=frac{n^2+n}{2}.
end{eqnarray}
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
$endgroup$
Here is my two cents:
begin{eqnarray}
M_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
*&*&*&*&cdots \
*&*&*&*& \
*&*&*&*& \
*&*&*&*& \
vdots&&&&ddots
end{pmatrix} hspace{.5cm} text{with $n^2$ elements}\ \ \
Skew_{n times n}(mathbb{R}) & text{has form} & begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix}
end{eqnarray}
For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $dim(Skew_{n times n}(mathbb{R}) + Sym_{n times n}(mathbb{R})) = dim(M_{n times n}(mathbb{R}))$ and $dim(Skew_{n times n}(mathbb{R}))=frac{n^2-n}{2}$ then we have that
begin{eqnarray}
frac{n^2-n}{2}+dim(Sym_{n times n}(mathbb{R})))=n^2
end{eqnarray}
or
begin{eqnarray}
dim(Sym_{n times n}(mathbb{R})))=frac{n^2+n}{2}.
end{eqnarray}
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
answered Apr 17 '13 at 1:53
TrancotTrancot
1,6901343
1,6901343
add a comment |
add a comment |
$begingroup$
This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n \ 2 end{array} right) = frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct symbols, where now a symbol is either an index ($1 , ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n+1 \ 2 end{array} right) = frac{n(n+1)}{2}$ such sets.
answered Jul 9 '14 at 19:37
StirlingStirling
36429
$endgroup$
add a comment |
$begingroup$
This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n \ 2 end{array} right) = frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct symbols, where now a symbol is either an index ($1 , ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n+1 \ 2 end{array} right) = frac{n(n+1)}{2}$ such sets.
answered Jul 9 '14 at 19:37
StirlingStirling
36429
$endgroup$
add a comment |
$begingroup$
This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n \ 2 end{array} right) = frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct symbols, where now a symbol is either an index ($1 , ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n+1 \ 2 end{array} right) = frac{n(n+1)}{2}$ such sets.
answered Jul 9 '14 at 19:37
StirlingStirling
36429
$endgroup$
This is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.
In order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n \ 2 end{array} right) = frac{n(n-1)}{2}$ such sets.
Similarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set ${i,j}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set ${i,j}$ of distinct symbols, where now a symbol is either an index ($1 , ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $left( begin{array}{@{}c@{}} n+1 \ 2 end{array} right) = frac{n(n+1)}{2}$ such sets.
answered Jul 9 '14 at 19:37
StirlingStirling
36429
answered Jul 9 '14 at 19:37
StirlingStirling
36429
answered Jul 9 '14 at 19:37
StirlingStirling
36429
answered Jul 9 '14 at 19:37
StirlingStirling
36429
36429
add a comment |
add a comment |
$begingroup$
The dimension of symmetric matrices is $frac{n(n+1)}2$ because they have one basis as the matrices ${M_{ij}}_{n ge i ge j ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is ${M_{ij}}_{n ge i > j ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
$endgroup$
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
add a comment |
$begingroup$
The dimension of symmetric matrices is $frac{n(n+1)}2$ because they have one basis as the matrices ${M_{ij}}_{n ge i ge j ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is ${M_{ij}}_{n ge i > j ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
$endgroup$
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
add a comment |
$begingroup$
The dimension of symmetric matrices is $frac{n(n+1)}2$ because they have one basis as the matrices ${M_{ij}}_{n ge i ge j ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is ${M_{ij}}_{n ge i > j ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
$endgroup$
The dimension of symmetric matrices is $frac{n(n+1)}2$ because they have one basis as the matrices ${M_{ij}}_{n ge i ge j ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is ${M_{ij}}_{n ge i > j ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.
Note that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
edited Aug 23 '12 at 10:06
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
answered Aug 23 '12 at 9:59
Rijul SainiRijul Saini
1,7991020
1,7991020
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
add a comment |
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
But this is no explanation why the symmetric matrices have the specified $dim$.
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:03
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Do you mean $n^2$?
$endgroup$
– enzotib
Aug 23 '12 at 10:04
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited.
$endgroup$
– Rijul Saini
Aug 23 '12 at 10:07
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
$begingroup$
@RijulSaini: Thanks, but enzotib's answer seems to be easier to understand!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:13
add a comment |
$begingroup$
There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$begin{matrix}
begin{pmatrix}
0 &*' \
*& 0 \
end{pmatrix} & *
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' \
*& 0 & *' \
* & * & 0 \
end{pmatrix} & begin{matrix}
*& \
* & *
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' &*'\
*& 0 & *' &*'\
* & * & 0 &*'\
* & * &* & 0
end{pmatrix} & begin{matrix}
* & & \
*&* & \
* & * &*
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix} & begin{matrix}
*& \
*&*& \
*&*&*& \
vdots&&&&ddots
end{matrix}
end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + dots + (n-1).$$
So by Gauss, we have $$frac{(n-1)(n-1+1)}{2} = frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$frac{(n-1)n}{2} + n = frac{n^2 + n}{2}$$
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
$endgroup$
add a comment |
$begingroup$
There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$begin{matrix}
begin{pmatrix}
0 &*' \
*& 0 \
end{pmatrix} & *
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' \
*& 0 & *' \
* & * & 0 \
end{pmatrix} & begin{matrix}
*& \
* & *
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' &*'\
*& 0 & *' &*'\
* & * & 0 &*'\
* & * &* & 0
end{pmatrix} & begin{matrix}
* & & \
*&* & \
* & * &*
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix} & begin{matrix}
*& \
*&*& \
*&*&*& \
vdots&&&&ddots
end{matrix}
end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + dots + (n-1).$$
So by Gauss, we have $$frac{(n-1)(n-1+1)}{2} = frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$frac{(n-1)n}{2} + n = frac{n^2 + n}{2}$$
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
$endgroup$
add a comment |
$begingroup$
There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$begin{matrix}
begin{pmatrix}
0 &*' \
*& 0 \
end{pmatrix} & *
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' \
*& 0 & *' \
* & * & 0 \
end{pmatrix} & begin{matrix}
*& \
* & *
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' &*'\
*& 0 & *' &*'\
* & * & 0 &*'\
* & * &* & 0
end{pmatrix} & begin{matrix}
* & & \
*&* & \
* & * &*
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix} & begin{matrix}
*& \
*&*& \
*&*&*& \
vdots&&&&ddots
end{matrix}
end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + dots + (n-1).$$
So by Gauss, we have $$frac{(n-1)(n-1+1)}{2} = frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$frac{(n-1)n}{2} + n = frac{n^2 + n}{2}$$
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
$endgroup$
There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$begin{matrix}
begin{pmatrix}
0 &*' \
*& 0 \
end{pmatrix} & *
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' \
*& 0 & *' \
* & * & 0 \
end{pmatrix} & begin{matrix}
*& \
* & *
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0 &*' & *' &*'\
*& 0 & *' &*'\
* & * & 0 &*'\
* & * &* & 0
end{pmatrix} & begin{matrix}
* & & \
*&* & \
* & * &*
end{matrix}
end{matrix}$$
$$begin{matrix}
begin{pmatrix}
0&*'&*'&*'&cdots \
*&0&*'&*'& \
*&*&0&*'& \
*&*&*&0& \
vdots&&&&ddots
end{pmatrix} & begin{matrix}
*& \
*&*& \
*&*&*& \
vdots&&&&ddots
end{matrix}
end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + dots + (n-1).$$
So by Gauss, we have $$frac{(n-1)(n-1+1)}{2} = frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$frac{(n-1)n}{2} + n = frac{n^2 + n}{2}$$
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
answered Aug 26 '17 at 11:56
IAmNoOneIAmNoOne
2,64541221
2,64541221
add a comment |
add a comment |
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$begingroup$
Do you mean symmetric (not normal) in the title? And do you mean the $dim$ of linear spaces of such matrices, not the $dim$ of the matrices, right?
$endgroup$
– enzotib
Aug 23 '12 at 9:59
$begingroup$
I did edit it - thanks for the reminder!
$endgroup$
– Christian Ivicevic
Aug 23 '12 at 10:02
$begingroup$
You did not edit it correctly; $mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:38
$begingroup$
And you should say that $mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true.
$endgroup$
– Marc van Leeuwen
Aug 23 '12 at 12:40