Linear Algebra - Rank of a matrix

A is a $100 \times 100$ matrix.

The element in the $i^{th}$ row and $j^{th}$ column is given by $i^2 + j^2$

Find the rank

Solution 1:

Hint: Show that each row is a linear combination of the vectors $(1,4,9,\ldots,100^2)$ and $(1,1,1,\ldots,1)$.

Solution 2:

Hint: rank of sum $\le$ sum of ranks.

Solution 3:

I think the rank is $2$.

Fact *: Rank of a matrix = Column rank.

Fact 1:When we perform any row operation or column operation to a matrix then the rank of the matrix does not change.

Now we perform the column operations: Subtract the first column from all the columns.

Then the matrix has all its $i$th ($2\le i\le 100$) column as $(i^2-1)(1,1,\dots,1)^t$ and the first column is $(1^2+1^2,2^2+1^2,\dots,100^2+1^2)^t$

Now all the columns of this matrix is linear combination of the following two vectors:$\{(1^2,2^2,3^2\dots,n^2)^t,(1,1,1,\dots,1)^t\}=S$

As all the columns of this reduced matrix is spanned by the vectors in $S$ and $S$ is a linearly independent set, the column rank of the reduced matrix is $2$. Implying by Fact 1 and Fact * that the rank of the matrix is $2$

Solution 4:

Hint:

Let $[B]_{ij} = i^2$, $[C]_{ij} = j^2$. Notice that $A=B+C$ (indeed, $C=B^T$).

What can you say about the rank of $B,C$?

Also, notice that $A$ contains the submatrix $\begin{bmatrix} 2 & 5 \\ 5 & 8\end{bmatrix}$, which is invertible.

If we must use Matlab (or Octave, in my case):

rank(ones(100,1)*(1:100).^2+(ones(100,1)*(1:100).^2)')