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If $A=[a_{ij}]$ is a skew-symmetric matrix, then write the value of $$ \sum_i \sum_j a_{ij}$$

My doubt is that what is the meaning of $ \sum_i \sum_j ?$ Is it the same as $\sum_{ij}?$
Please offer your assistance.
Thank you

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Yes, sum over all pairs $(i.j)$ for $1 \le i \le n$ and $1 \le j \le n$ where $n$ is the dimension of the matrix. –  coffeemath May 24 '13 at 12:16

1 Answer 1

up vote 2 down vote accepted

As for the notaion. It is easier to understand on concrete example. Say $n=3$, then $$ \begin{align} \sum\limits_{i=1}^3\sum\limits_{j=1}^3a_{ij} &=\sum\limits_{i=1}^3(a_{i1}+a_{i2}+a_{i3})\\ &=(a_{11}+a_{12}+a_{13})+(a_{21}+a_{22}+a_{23})+(a_{31}+a_{32}+a_{33})\\ &=(a_{11}+a_{12}+a_{13}+a_{21}+a_{22}+a_{23}+a_{31}+a_{32}+a_{33})\\ &=\sum\limits_{i,j=1}^3 a_{ij} \end{align} $$

As for the original problem. Elements of the diagonal are zeros (why?). So you need to perform calculations only for nondiagonal entries. Divide them into pairs $a_{ij}+a_{ji}$, recall definition of skew symmetric matrix and conclude...

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okay, i know that. so are they one and the same? –  chndn May 24 '13 at 12:22
    
all pairs gives the same sum –  no identity May 24 '13 at 12:22
    
no that si not what i asked is $\Sigma_i \Sigma_j [aij] = \Sigma_{i j} [aij]$? –  chndn May 24 '13 at 12:23
2  
yes it is $\phantom{}$ –  no identity May 24 '13 at 12:24
    
thanks ............. +1 –  chndn May 24 '13 at 12:36

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