Finding the rank of a matrix $A$ Find the rank of the $n \times n$ matrix $A = [i + j]_{i,j \le n}$ (over $C$).
C here should be the complex space; although i am having trouble interpreting what A exactly is, I do not understand the middle bracket notation too well, is it ixn+jxn matrix? Any hint on this problem would be apprecaited. Thanks.
 A: This means the matrix with entries $a_{ij} = i+j$.
Hint: write $A$ as the sum of two simpler matrices.
A: The first row is
$$R_1=\begin{bmatrix}2 & 3 & \cdots & n+1 \end{bmatrix}$$
the second one
$$R_2=\begin{bmatrix}3 & 4 & \cdots & n+2 \end{bmatrix}$$
the difference is
$$R_2-R_1=\begin{bmatrix}1 & 1 & \cdots & 1 \end{bmatrix}$$
and so on by row operation any other row is a combination of this first two
$$ \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_2-R_1}\to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_3-(R_1+2R_2)}\to$$
$$  \to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix} \stackrel{R_4-(R_1+3R_2)}\to \ldots \stackrel{R_n-(R_1+(n-1)R_2)}\to \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 1 & 1 & \cdots & 1 \\0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\end{bmatrix} $$
A: $A$ is an $n \times n$ matrix whose $(i,j)$-th entry is $A_{i,j} = i+j$. So $$A = \begin{bmatrix}2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\ \vdots & \vdots & \ddots & \vdots \\ n+1 & n+2 & \cdots & 2n\end{bmatrix}.$$
As a hint for finding the rank of $A$, note that $$A = \begin{bmatrix}1 & 1 & \cdots & 1 \\ 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \cdots & n\end{bmatrix} + \begin{bmatrix}1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \cdots & n\end{bmatrix}.$$ Can you use this to find a basis for the column space of $A$? How many elements does that basis have?
