Square of Ones Matrix

Theorem
Let $$\mathbf{J} = \left[{1}\right]_n$$ be a square ones matrix of order $n$.

Then $$\mathbf{J}^2 = n \mathbf{J}$$.

Proof
Follows directly from the definition of matrix multiplication:


 * $$\forall i \in \left[{1 \, . \, . \, m}\right], j \in \left[{1 \, . \, . \, p}\right]: c_{i j} = \sum_{k=1}^n a_{i k} \circ b_{k j}$$

In this case, $$m = n$$ and $$a_{ik} = b_{kj} = 1$$.

Hence $$c_{i j} = \sum_{k=1}^n 1 \times 1 = n$$.