For a skew-symmetric matrix $A$, it satisfies the property [1]:

$$A^T = -A$$

Now, let's show that all the diagonal elements of a skew-symmetric matrix are zero.

Consider a general skew-symmetric matrix $A$ with the size of $n \times n$. Denote its elements as $a_{ij}$, where $i$ and $j$ represent the row and column indices, respectively. The diagonal elements of the matrix $A$ are those where $i = j$, so $a_{ii}$ represents the diagonal element.

Now, we have the property that $A^T = -A$. This means that the transpose of matrix $A$ is equal to its negative. Let's consider the diagonal elements:
$$(A^T)_{ii} = (-A)_{ii}$$

By definition, $(A^T)_{ii}$ represents the element in the $i$-th row and $i$-th column of the transpose of $A$, and $(-A)_{ii}$ represents the element in the $i$-th row and $i$-th column of the negation of $A$.

For the transpose, $(A^T)_{ii} = a_{ji}$ (where $j = i$)

For the negation, $(-A)_{ii} = -a_{ii}$

Now, we have:

$a_{ji} = -a_{ii}$

Since the transpose of a matrix simply swaps rows and columns, the $i$-th row of $A$ becomes the $i$-th column of $A^T$, and the $j$-th column of $A$ becomes the $j$-th row of $A^T$. Thus, $a_{ji}$ is equal to $a_{ii}$, which is the diagonal element of $A$.

So, we have:

$a_{ii} = -a_{ii}$

This implies that $a_{ii}$ must be equal to its negation, which is only possible if $a_{ii} = 0$. Therefore, in a skew-symmetric matrix, all the diagonal elements must be equal to $0$.