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Show that diagonal entries of a skew symmetric matrix are zero.

 
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Feb2-11, 08:09 AM   #1
 

Show that diagonal entries of a skew symmetric matrix are zero.

I'm pretty inexperienced in proof writing. So not sure if this was valid.

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.

Is this good enough?

Thanks.
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Feb2-11, 09:42 AM   #2
 
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I think it would work as a valid proof.
Nov20-12, 06:53 PM   #3
 
Quote by inknit View Post
I'm pretty inexperienced in proof writing. So not sure if this was valid.

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.

Is this good enough?

Thanks.
It's great. Thanks!
Nov20-12, 07:29 PM   #4
 
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Show that diagonal entries of a skew symmetric matrix are zero.

Quote by inknit View Post
I'm pretty inexperienced in proof writing. So not sure if this was valid.

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.
A touch better than just saying that would be to note that if a(j,j)= -a(j, j) then (adding a(j,j) to both sides) 2a(j,j)= 0 so (dividing both sides by 2) a(j,j)= 0.

So therefore all the diagonal entries of a skew symmetric matrix are 0.

Is this good enough?

Thanks.
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