TITLE: Rank/determinant of the $n\times n$ matrix $((a_{ij}))$ where $a_{ij}=(i+j-1)^2$
QUESTION [5 upvotes]: I am trying to find the rank and determinant of the following $n\times n$ matrix : 
$$A=\begin{bmatrix}1^2&2^2&3^2&\cdots&n^2
   \\ 2^2&3^2&4^2&\cdots&(n+1)^2
   \\\vdots&\vdots&\vdots&\ddots&\vdots\\n^2&(n+1)^2&(n+2)^2&\cdots&(2n-1)^2
   \end{bmatrix}$$
I verified that the determinant of $A$ vanishes for $n> 3$, implying that $A$ has full rank for $n\le 3$. And the rank seems to remain $3$ for $n>3$. But I could not provide a rigorous proof.
Is there a way to deduce the rank/determinant without reducing $A$ to echelon form or any other shortcut? Or how can I easily find the number of linearly independent rows/columns? Any hints would be great.
Related question: Determinant of the matrix with $a_{i,j} = (i+j)^2$.

REPLY [2 votes]: The fact that the upper left $3 \times 3$ submatrix of $A$ has nonzero determinant implies that the first three columns of $A$ are linearly independent, implying that the rank of $A$ is at least 3.
On the other hand, the column space of $A$ is contained in the image of the linear map which takes a quadratic polynomial $p \in P_2(\mathbb{R})$ to $(p(1), p(2), \ldots, p(n))$.  Since $P_2(\mathbb{R})$ has dimension 3, this implies that the rank of $A$ is at most 3.