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This has already been asked but I can't seem to understand why its $\frac{1}{2}n(n+1)$ elements for the basis of symmetric $n\times n$ matrices and $\frac{1}{2}n(n-1)$ for skew symmetric $n\times n$ matrices. Any mechanical explanation that I can easily see?

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Basically a symmetric matrix is determined by its coefficients on the lower triangle (including the diagonal) because then the others are determined by symmetry. So you have $n(n+1)/2$ parameters.

For an anti-symmetric matrix, it's almost the same except that everything on the diagonal must be $0$. So you only have $n(n+1)/2-n=n(n-1)/2$ parameters.

To see these numbers, divide your square in $3$ parts : the strict lower triangle, the strict upper triangle and the diagonal. Obviously, the diagonal has $n$ spots. Now the rest ($n^2-n$ spots) is equally divided in the strict lower and strict upper triangles, so each have $(n^2-n)/2$ spots, ie $n(n-1)/2$. So the large lower triangle is $n(n-1)/2+n = n(n+1)/2$.

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  • $\begingroup$ Why is it $n(n+1)/2$ for the lower triangle? Doesn't the triangle have n entries in the non hypotenuse sides? Shouldn't it be $n*n/2$? $\endgroup$ – Omrane Mar 27 '16 at 20:02
  • $\begingroup$ I edited my answer to make it clear. $\endgroup$ – Captain Lama Mar 27 '16 at 20:06
  • $\begingroup$ I see! That is so much clearer. One question, shouldn't the diagonal have $\sqrt{2} n$ spots since we are in a n x n square? $\endgroup$ – Omrane Mar 27 '16 at 20:09
  • $\begingroup$ No, its length is $\sqrt{2}n$, but if you divide your $n\times n$ square in small $1\times 1$ blocks, there will be $n$ blocks in the diagonal, each of diagonal length $\sqrt{2}$ (so the total diagonal length is indeed $\sqrt{2}n$). $\endgroup$ – Captain Lama Mar 27 '16 at 20:11
  • $\begingroup$ I see! Took the example for n=1,2,3,... and that seems logical now. Thanks a bunch for your answer. For anti-symmetric, why must the diagonal be 0? It only says that $a_{ij}=-a_{ji}$. $\endgroup$ – Omrane Mar 27 '16 at 20:13
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For symmetric matrices entries above the diagonal are same as entries below the diagonal. So you just need to find how many entries are there above the diagonal including diagonal entries. For that subtract the number of entries below the diagonal from $n^2$(total number of entries) to get your answer.

Number of entries below the first diagonal entry = $n-1$

Number of entries below the second diagonal entry = $n-2$

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Number of entries below the $n^{th}$(last) diagonal entry = $0$

Thus, number of entries below diagonal= $(n-1)+(n-2)+\cdots+1+0=\frac{n(n-1)}{2}$

Now, $n^2-\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$

which is the required answer.

For Skew- symmetric case diagonal entries are all zero and entries below diagonal are just the negative of entries above diagonal. So only entries below the diagonal are enough to get the whole matrices, which is determined by $\frac{n(n-1)}{2}$(calculated above ).

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  • $\begingroup$ What do you mean by "first diagonal entry"? $\endgroup$ – Omrane Mar 27 '16 at 20:05
  • $\begingroup$ $a_{11}$, basically entry at intersection of first row and first column. $\endgroup$ – User Mar 27 '16 at 20:07

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