in a skew symmetric, how do we explain that its diagonal entries has to be all 0 without giving examples to support it
For the definition of a skew-symmetirc matrix, see
http://en.wikipedia.org/wiki/Skew-symmetric_matrix
The requirement for such a matrix is: aij = -aji
For the diagonal elememnts aii, clearly the requirement is that they be zero. A number can only equal its negative if it is zero.
lim x infinitive (4x-1/2x+9)
In order to explain why the diagonal entries of a skew symmetric matrix are always zero, we can approach it mathematically.
Let's consider a general skew symmetric matrix A:
A = [a11 a12 a13]
[a21 a22 a23]
[a31 a32 a33]
To say that A is skew symmetric means that it satisfies the condition A = -A^T, where A^T is the transpose of A.
The transpose of A is obtained by switching its rows and columns:
A^T = [a11 a21 a31]
[a12 a22 a32]
[a13 a23 a33]
Now, if we negate this transpose and consider -A^T, we get:
-A^T = [-a11 -a21 -a31]
[-a12 -a22 -a32]
[-a13 -a23 -a33]
For A to satisfy the condition A = -A^T, we compare the corresponding elements of A and -A^T. Since a matrix is equal to its transpose only when all the corresponding elements are the same, we have:
a11 = -a11
a22 = -a22
a33 = -a33
From these equations, we can conclude that a11, a22, and a33 must be equal to their negatives, which implies they must be equal to zero:
a11 = 0
a22 = 0
a33 = 0
Therefore, the diagonal entries of a skew symmetric matrix are always zero, and this can be mathematically deduced without the need for specific examples.