Square root of a square matrix of ones Let $J_n$ be the square matrix of ones (all entries are a one). I want to characterize the "square root" of $J_n$ where the square root is a matrix $A$ such that $A'A=J_n$ and $A$ is $k\times n$.
In particular, I think that any such matrix $A$ needs to have all columns be the same. For instance, if $n=2$ and $A = [\mathbf{a} ~ \mathbf{b}]$, then $\mathbf{a} = \mathbf{b}$. If we assume that $n=2$ and $k=2$ ($A$ is $2\times 2$), this can be shown by solving quadratic equations. But I have not been able to show that this is a more general fact of such square root matrices. Is this a general property for any $n$ and $k$?
 A: The answer is yes.
Suppose that $A'A = J_n$. Note that the row-space of $A$ is equal to the row-space of $A'A = J_n$, which is the span of the vector $(1,1,\dots,1)$. Thus, every row of $A$ is a multiple of $(1,1,\dots,1)$. From this, we can conclude that the columns of $A$ must be equal.
A: All columns of $A$ must be unit vectors $\mathbf a_i$ and all dot products of those vectors must be 1, i.e. $\mathbf a_i \cdot \mathbf a_j = 1$. One can show that all columns $\mathbf a_i$ must be identical. Indeed, $\mathbf a_i \cdot \mathbf a_j \leq \frac{|\mathbf a_i|^2 + |\mathbf a_j|^2}{2} = 1$, and the equality is only possible when $\mathbf a_i = \mathbf a_j$.
A: Start with:
$$\rm{rank}\ J=rank\ A^{'}A \le min(rank\ A^{'}, rank\ A)$$
From this it is clear that:
$$\rm{rank}\ J=rank\ A=1$$
This shows that the columns $\textbf{a}_i$ are multiples of each other.
Calculating the value of the top left entry of the product $A^{'}A$ leads to:
$$ |\textbf{a}_1|^2=1 $$
And similar calculations along the diagonal show that the other columns must also be unit vectors.
Thus, all columns must be the same unit vector.
