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Let $A_n$ be the real $n × n$ matrix $(n ≥ 2)$ whose $(i, j)$ entry is $i − j$. What is the rank of $A_n$ as a function of $n$?

My attempt:- $A_2$ has rank $2$. It is obvious. $A_3$ has also rank $2$. Since, $A_3$ has determinant $0$ and has a submatrix of order $2$ with determinant non-zero. Similarly, I could conclude that $A_4$ has rank $2$. Using this method I can't go beyond. I also know that $A_n$ is a skew-symmetric matrix. $\det A_n=0,\forall n.$ I am not able to draw conclusion beyond this. Please help me.

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    $\begingroup$ Unless I'm misreading something, if $\det(A_3)=2$, then $A_3$ is invertible and so has full rank, i.e. 3. $\endgroup$ – Reveillark Nov 6 '19 at 2:34
  • $\begingroup$ OP has a typo (actually multiple). He said at the end that $det(A_n) = 0$, which is true for $n$ odd due to skew symmetrty. (actually, holds for all $n\neq 2$ of this form). $\endgroup$ – Calvin Lin Nov 6 '19 at 2:38
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    $\begingroup$ @CalvinLin: Please don't assume that a user must be male. There is a lot of discrimination against women in mathematics; defaulting to male pronouns perpetuates that bias. $\endgroup$ – Greg Martin Nov 6 '19 at 2:43
  • $\begingroup$ @Reveillark Thank you for pointed out my mistake. $\endgroup$ – Truth_searcher Nov 6 '19 at 4:56
  • $\begingroup$ @GregMarting You are probably refering to the previous century, or in any case many, many years ago. $\endgroup$ – user26857 Nov 6 '19 at 5:47
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Hint: Can you show that the matrix $B = \begin{bmatrix}1 & 1 & \cdots & 1 \\ 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \cdots & n\end{bmatrix}$ (i.e. $B_{i,j} = i$) has rank $1$?

Similarly, can you show that the matrix $C = \begin{bmatrix}1 & 2 & \cdots & n \\ 1 & 2 & \cdots & n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \cdots & n\end{bmatrix}$ (i.e. $C_{i,j} = j$) has rank $1$?

Once you do that, what can you say about the rank of $A = B-C$?

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    $\begingroup$ This is good. In different words, the OP should prove that the row space of $A_n$ is spanned by $(1,1,1,\ldots,1)$ and $(1,2,3,\ldots,n)$. $\endgroup$ – Jyrki Lahtonen Nov 6 '19 at 4:38
  • $\begingroup$ I could prove $B,C$ has rank $1$. $\endgroup$ – Truth_searcher Nov 6 '19 at 5:03
  • $\begingroup$ rank($A$) $\leq $ rank($B$)+rank($-C$)=$2$ $\endgroup$ – Truth_searcher Nov 6 '19 at 5:07
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    $\begingroup$ Also, It has a submatrix of order 2 with non zero determinants. Hence Rank(A)$\ge 2$ $\endgroup$ – Truth_searcher Nov 6 '19 at 5:08
  • $\begingroup$ @JyrkiLahtonen That's the gist of my solution. $\endgroup$ – Calvin Lin Nov 6 '19 at 5:08
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Hint: Show that column $k$ of the matrix is given by $A + k B$. (Find these column vectors)

Hence, the matrices have rank 2.

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  • $\begingroup$ I don't understand your hint. Can you please expand it bit. $\endgroup$ – Truth_searcher Nov 6 '19 at 5:02
  • $\begingroup$ @Truth_searcher Assuming the hint, do you see why this tells us that the matrix has rank 2 if $B$ is not a multiple of $A$, rank 1 if $B$ is a multiple of $A$ and neither are 0, and rank 0 if $A = B = 0$? $\endgroup$ – Calvin Lin Nov 6 '19 at 5:07
  • $\begingroup$ Okay. Thank you very much. $\endgroup$ – Truth_searcher Nov 6 '19 at 5:09
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The upper left $2\times 2$ matrix is always the same and has rank $2$, so $\operatorname{rank}A_n\ge 2$ for all $n\ge 2$. Moreover, $$ a_{1j}-a_{ij} = 1-j-(i-j) = 1-i $$ means that subtracting row $i$ from row $1$ is always a multiple of the vector $(1,1,\ldots,1)$ and thus shows that $\operatorname{rank}A_n = 2$ for all $n$.

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  • $\begingroup$ I understood your argument for $A_n\ge 2$. I don't understand the other way argument. $\endgroup$ – Truth_searcher Nov 6 '19 at 5:00
  • $\begingroup$ @Truth_searcher It means subtracting row $i$ from row $1$ is always a multiple of the vector $(1,1,\ldots,1)$. I edited. $\endgroup$ – amsmath Nov 6 '19 at 5:09

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