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

[inknit]

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.

 

[rock.freak667]

I think it would work as a valid proof.

 

[John Mark]

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!

 

[HallsofIvy]

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.