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In a linear algebra book, I found the following problem.

Find the rank of the matrix

$$\begin{bmatrix} 1 & x & x & \dots & x\\ x & 1 & x & \dots & x\\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ x & x & x & \dots & 1\end{bmatrix}$$

I used four pages of my scrap notebook to find some patterns for the $3 \times 3$ and $4 \times 4$ cases, where the matrix is reduced to triangular ones, and maybe prove by induction.

As I need to read ahead, I stopped spending time on this. Besides, I have a job. I guess there got to be a simple way to solve this problem. Does anyone have an idea?

Suggestion: As this is known to be duplicate, I might like to edit. Many solution to this problem I see in stackexchange uses the determinant of the given matix. Unfortunately, the problem appears before the section where the determinant is defined in a book I read (This is not a lie.)

Thus, I would like to know a solution which doesn't use determinant or eigenvalues or orthogonal space of {1, 1, ..., 1}.

Community
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Grown pains
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  • What do you want exactly? Rank or determinant? – Error 404 Jun 25 '17 at 04:24
  • Oops. Please show me a nice argument to find the rank of the given matrix. – Grown pains Jun 25 '17 at 04:32
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    "I have a job" is the least motivating line you could have used, I think. "I have a job so I don't have time to waste on these trivialities, while you, on the other hand..." Why do you think we do not have jobs? – Mariano Suárez-Álvarez Jun 25 '17 at 05:00
  • @Mariano This may not be a trivial one to me since I actually can't solve it. It's just I am tired of facing up to this problem. I am sure there is a clever guy who has a job and can solve instantaneously. – Grown pains Jun 25 '17 at 06:53
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    Well, I was suggesting you edit your question in order to be a bit less obnoxious... In any case: why are you in a hurry to know the rank of this matrix? If you are tired now and you have other more pressing things to do, then rest, do those things, and then spend some more time on the problem. – Mariano Suárez-Álvarez Jun 25 '17 at 06:57

2 Answers2

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Let $\bf e$ be the all $1$ colomn vector $[1,\ldots,1]^T$, and let $F={\bf e}^\bot$ which is the $(n-1)$-dimensional vector subspace orthogonal to $\bf e$, $$F=\{[u_1,\ldots,u_n]^T:u_1+\cdots+u_n=0\}$$ Finally, let the given matrix be denoted by $M$.

Clearly $M{\bf e}=(1+(n-1)x){\bf e}$ and $M{\bf u}=(1-x){\bf u}$ for every ${\bf u}\in F$. Thus, we have the following cases:

  • If $x\notin\{1,-\frac{1}{n-1}\}$ then the matrix is invertible that is ${\rm rank}\,M= n$, moreover $\det M=(1-x)^{n-1}(1+(n-1)x)$.
  • If $x=1$ then ${\rm ker}\, M= F$ and ${\rm rank}\,M= 1$.
  • If $x =-\frac{1}{n-1}$ then ${\rm ker}\, M= \mathbb{R} {\bf e}$ and ${\rm rank}\,M= n-1$.
Rodrigo de Azevedo
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Felix Klein
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  • A nice solution which I can hardly think up in less than an hour. However, the notion of breaking $\mathbb{R}^{n}$ into two subspaces has not appeared yet in my reading. – Grown pains Jun 25 '17 at 06:46
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Hint: Since the given matrix is symetric, Rank equals number of non zero eigen values.

ASB
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  • People are using some property of eigenvalues. But this problem appears before eigenvalues are discussed in the book. Thus, I want to ask for a solution without such notions. – Grown pains Jun 25 '17 at 06:38