Intuitively Finding Rank of a Matrix Let M be a square matrix, where $M[i,j] = (i-j)^2$.
I am told to find its rank.
The solution says that it is easy to see that the Columns of M are spanned by $\{ [1, 1, 1, 1,\dots, 1]^T , [1, 2,\dots, n]^T, [1, 2^2,\dots, n^2]^T \}$ .`
So, $rank(M) = 3$.
Well Rank = max. no of linearly independent columns. Why is the first column of $[1, 1,\dots, 1]$ required?  

A random column of $M$ looks like $(i-j)^2$, where $j$ is fixed. Note that $(i-j)^2 = i^2 + j^2 - 2ij$.
We want to express $C_j$, the $j$th column of $M$, as the linear combination of $ A= [1,...,1]^T ,$ $ B = [1,...,n]^T $ and $D = [1,4,...,n^2]^T$. From the above relation of $(i,j)$, we have that $C_j = D -2jB + j^2A$. This is because $A_i = 1, B_i = i,D_i = i^2$, so we see that $C_{ij} = i^2 - 2ij + j^2$, which is true.
Hence, the rank of such a matrix must be $3$.
