Every column of $A$ is identical $\implies A$ has only one linearly independent column $\implies$ rank of $A = 1$.
$\therefore$ option (A) is not True.
$\begin{align}A \times A = A^2 \implies A_{ij} &= row_i(A) . column_j(A) = [(x_i) \text{ } (x_i) … (x_i)] \times [1 \text{ } 3 … (2n-1) \text{ } (-n^2)] ^ T \\ &= (x_i)*1 + (x_i)*3 + … + (x_i)*(2n-1)+(x_i)*(-n^2) \\ &=(x_i)(1 + 3 + … + (2n-1)+(-n^2)) \\ &=(x_i)(\frac{n}{2} * (2*1 + (n-1)*2) - n^2) \\ &=(x_i)(n^2-n^2) \\ &= 0 \end{align}$
where $x_i$ is an element occupying $row_i(A)$ as every element in a $row$ are same.
$\implies$ rank of $A^2 = 0$
$\therefore$ option (B) is not True.
Since, A does not have a full rank, $det(A) = 0 =$ product of eigenvalues $\implies$ at least one eigenvalue has to be zero.
Consider, $(A-0 I)x = Ax = 0$
We know rank of $A=1$ and nullity of $A=n \implies$ There are $n$ linearly independent solutions of $Ax=0$.
$\therefore A$ has $n$ linearly independent eigenvectors corresponding to $0$.
$\implies$ Algebraic Multiplicity of $0$ must be equal to $n$ or $n+1$.
$\therefore A$ has atleast $n$ eigenvalues are $0$ ie atleast $n$ eigenvalues are repeating.
$\therefore$ option (C) is not True.
Trace of $A = 1 + 3 + … + (2n-1) + (-n^2) = 0$.
Since, we know $n$ eigenvalues are $0$, $(n+1)^{th}$ eigenvalue must also be $0$.
$\therefore$ option (D) is True.
Answer :- D.