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Given $n\geq 2$, let $M$ be a matrix with entries $1, \dots, n^2$, e.g.,

$$\begin{alignat*}{1} A_3 = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} & \qquad A_4 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & 16 \end{pmatrix} \end{alignat*}$$

How do I prove that $\operatorname{rank}(M)=2$? I am supposed to use only the elementary row operations. For me, the definition of $\operatorname{rank}$ is number of non-zero rows in a row-echelon form.

For a specific $n$, I can actually perform the row operations and obtain the row-echelon form. However, I don't know how to prove it in general. Should I prove it using induction? How do I proceed with the induction step? The matrices $A_n$ and $A_{n+1}$ are completely different!

By the way, is there any specific name for this type of matrices?

Rodrigo de Azevedo
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Mike V.D.C.
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2 Answers2

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Notice that the columns of $A_n$ obey this property: $$\vec v_n=2\vec v_{n-1} - \vec v_{n-2}$$

This means that only the first two columns are linearly independent, and everything after that can be represented as a linear combination of the the first two columns. Thus, $\rm Col(A)=2=Rank(A)$

Another solution using row reduction:

$R_2-R_1$ of every matrix is a row of $4$'s.

We can replace $R_2$ with $\frac{1}{4}(R_2-R_1)$ to get a row of $1's$. We can then replace every $R_n$ with $n \ge 3$ with $R_n-R_1-kR_2$ where $k$ is the number of $1$'s that need to be subtracted from $R_2-R_1$ to get a row of $0$'s.

Thus, we get the first row of $1,2,3 \ldots$ and the second row of just ones, rows that can't be turned into zeroes, meaning that $\rm Rank(A)=2$.

I do not know what these matrices are called.

D.R.
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  • Thanks @D.R., but I am supposed to prove it only using elementary row operations / REF / RREF. – Mike V.D.C. Aug 06 '17 at 05:08
  • I added a solution with RREF – D.R. Aug 06 '17 at 05:39
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    Thanks @D.R. Now the proof is clear to me! By the in your arguments, whenever you say 4, I presume its $n$. e.g. $R_2-R_1$ is a row of $n$'s. – Mike V.D.C. Aug 06 '17 at 06:04
  • Yes, that is correct! – D.R. Aug 06 '17 at 15:39
  • The first solution only proves that the rank is $\le 2$. You need a similar observation to the one you did in the second solution. – jjagmath Jul 01 '23 at 11:25
  • @jjagmath ah, sure. Thanks for the correction. Though I think one can just say that the 2nd column is not a scalar multiple of the 1st column, since comparing the first element the scalar would have to be $2$, and comparing the last element the scalar would have to be $<2$. – D.R. Jul 01 '23 at 18:38
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Given $n \geq 2$, let ${\bf r}_n := \begin{bmatrix} 1 & 2 & \cdots & n \end{bmatrix}^\top$ and

$$ {\bf A}_n := {\bf 1}_n {\bf r}_n^\top + n \left( {\bf r}_n - {\bf 1}_n \right) {\bf 1}_n^\top = \underbrace{\begin{bmatrix} | & | \\ {\bf 1}_n & {\bf r}_n \\ | & | \end{bmatrix}}_{=: {\bf V}_n} \underbrace{\begin{bmatrix} -n & 1 \\ n & 0 \end{bmatrix}}_{=: {\bf M}_n} \begin{bmatrix} | & | \\ {\bf 1}_n & {\bf r}_n \\ | & | \end{bmatrix}^\top = {\bf V}_n {\bf M}_n {\bf V}_n^\top $$

Thus, $\operatorname{rank} \left( {\bf A}_n \right) \leq 2$. Since $\operatorname{rank} \left( {\bf V}_n \right) = 2$ and $\operatorname{rank} \left( {\bf M}_n \right) = 2$, applying the Sylvester rank inequality twice, we conclude that $\operatorname{rank} \left( {\bf A}_n \right) \geq 2$. Thus, $\color{blue}{\operatorname{rank} \left( {\bf A}_n \right) = 2}$.

Code

from sympy import *

def A(n):
    r = Matrix([range(1,n+1)]).T
    V = Matrix([[ ones(n,1), r ]])
    M = Matrix([[ -n,  1 ],
                [  n,  0 ]])
    return V * M * V.T


print( (A(5), A(5).rank()) )

outputs

(Matrix([[ 1,  2,  3,  4,  5],
         [ 6,  7,  8,  9, 10],
         [11, 12, 13, 14, 15],
         [16, 17, 18, 19, 20],
         [21, 22, 23, 24, 25]]), 2)

Related

Rodrigo de Azevedo
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