# Find the rank of $A_n$ in terms of $n$

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.

• Unless I'm misreading something, if $\det(A_3)=2$, then $A_3$ is invertible and so has full rank, i.e. 3. – Reveillark Nov 6 '19 at 2:34
• 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). – Calvin Lin Nov 6 '19 at 2:38
• @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. – Greg Martin Nov 6 '19 at 2:43
• @Reveillark Thank you for pointed out my mistake. – Truth_searcher Nov 6 '19 at 4:56
• @GregMarting You are probably refering to the previous century, or in any case many, many years ago. – user26857 Nov 6 '19 at 5:47

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 $$A = B-C$$?

• 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 $(1,2,3,\ldots,n)$. – Jyrki Lahtonen Nov 6 '19 at 4:38
• I could prove $B,C$ has rank $1$. – Truth_searcher Nov 6 '19 at 5:03
• rank($A$) $\leq$ rank($B$)+rank($-C$)=$2$ – Truth_searcher Nov 6 '19 at 5:07
• Also, It has a submatrix of order 2 with non zero determinants. Hence Rank(A)$\ge 2$ – Truth_searcher Nov 6 '19 at 5:08
• @JyrkiLahtonen That's the gist of my solution. – Calvin Lin Nov 6 '19 at 5:08

Hint: Show that column $$k$$ of the matrix is given by $$A + k B$$. (Find these column vectors)

Hence, the matrices have rank 2.

• I don't understand your hint. Can you please expand it bit. – Truth_searcher Nov 6 '19 at 5:02
• @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 $A = B = 0$? – Calvin Lin Nov 6 '19 at 5:07
• Okay. Thank you very much. – Truth_searcher Nov 6 '19 at 5:09

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 $$n$$.

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