prove that the diagonal elements of a scew symmetric matrix are all zero

We know that for a skew symmetric matrix :-
aij = -aji.

Consider a 3x3 matrix 
according to the condition :-

Since , aij= -aji

a11 = -a11 

implies a11+a11 = 0 ;

2a11 =0 , hence a11 = 0;


a12 =0 and a13 =0.
Hope it helps :) 

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If a matrix is skew symmetric then A^T = - A, that is the transpose of A is equal to negative A.

This implies that if A = a(i,j), then a(j,i) = -a(i,j). If we're referring to diagonal entries, we can say a(j,j) = -a(j,j). The only way for this to be true is if a(j,j) = 0. So therefore all the diagonal entries of a skew symmetric matrix are 0.

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 For 1 <= i, j <= n let a_ij denote the row i, column j entry of a nxn matrix A. To say that A is skew symmetric is to say that A is equal to -A^t (minus the transpose of A). If you look in each entry this is saying that for all 1 <= i, j <= n one has 

a_ij = [the row i, column j entry of A] 
= [the row i, column j entry of -A^t] 
= -[the row i, column j entry of A^t] 
= -[the row j, column i entry of A] 
= -a_ji. 

In particular, taking j = i we see that for all 1 <= i <= n one has 

a_ii = -a_ii. 

Adding a_ii to both sides we deduce 2 a_ii = 0 for all 1 <= i <= n and hence a_ii = 0 for all 1 <= i <= n. So every diagonal entry of A is 0.
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Required answer.
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I got this ans from refrence book ao i thought some ome get help with this......

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Hope this helps

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