# Linear algebra rank question

Suppose $A$ is a square matrix of order $n \geq 4$, and $a_{ij} = i+j$ represents the entry in row $i$ and column $j$. What is the rank of $A$? So we have by the rank-nullity theorem,

$$n = \text{rank}(A) + \text{nullity}(A)$$

For example, for $n=4$, we have $$A = \begin{bmatrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 5 & 6 & 7 & 8 \end{bmatrix}$$

So it seems that every row shares elements with the other rows. Can we use this to find the nullity of $A$ and hence the rank of $A$?

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Subtracting the penultimate row from the last row, the antepenultimate row from the penultimate row, the preantepenultimate row from the antepenultimate row, and so on until we subtract the first row from the second row, our matrix is seen to be row-equivalent to $$\left(\begin{array}{cccc} 2&3&\cdots&n+1\\ 1&1&\cdots &1\\ 1&1&\cdots &1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{array}\right).$$ Subtracting the second row from all other rows we get $$\left(\begin{array}{cccc} 1 & 2 & \cdots &n\\ 1 & 1 & \cdots &1\\ 0 & 0 & \cdots &0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & \cdots & 0 \end{array}\right).$$ This matrix has rank $2$ (first and second rows are linearly independent), and nullity $n-2$. Since this matrix is row equivalent to our original matrix, it follows that the original matrix also has rank $2$ and nullity $n-2$.

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So you used elementary row operations to transform the matrix? –  Damien Jul 18 '12 at 16:51
@Damien: Yes: subtracting one row from another is an elementary row operation (it is equivalent to adding a [negative] multiple of one row to another. –  Arturo Magidin Jul 18 '12 at 16:57

The rank is 2. Subtract the first row from all the other rows; those rows become (1 1 1 ... 1), (2 2 2 ... 2), etc. Now clear rows 2 and further down using row 2. You're left with a matrix whose first row is (2 3 4 ... n+1) and second row is (1 1 1 ... 1) and zeroes everywhere else. The first two rows are linearly independent and hence the rank is 2.

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Subtract the first row from all other rows. You will get a matrix $B=(b_{i,j})$ with $$b_{i,j}=\left\{ \begin{array}{cc} j+1 & i=1\\ i & i>1\end{array} \right.$$

Can you continue from here?

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Let's write down the rows of an $n\times n$ matrix. What is the first row? $$r_1 = [1\quad 2\quad 3\quad \cdots \quad n+1]$$ What's the second row? That's a row with each entry $+1$ $$r_2 = [1+1\quad 1+2\quad 1+3\quad \cdots \quad 1+(n+1)]$$ Same for the 3rd row. Now it's evident that row $i$ is the linear combination $$1r_1 + (i-1)[1\quad 1\quad 1\quad \cdots \quad 1]$$ Now to show that the matrix has rank $2,$ it's enough to show that $r_1$ and $[1\quad \dots \quad 1]$ are linearly independent. One way is to check the rank of the matrix $$\pmatrix{1& 1& 1& \cdots & 1\\1& 2& 3& \cdots & n+1}$$ which is $2$ by trivial elimination.

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