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Let's call a sequence semi-arithmetic if the difference of any two consecutive elements of it can take two different values only.(E.g the sequence $3,7,11,12,16,17,18$ is semi-arithmetic). For the matrix $A$ it holds that the first three rows of it form a semi-arithmetic sequence and all the other rows of it form an arithmetic sequence. Show that in this case $r(a) \le 9$ where $r$ denotes rank.

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I wrote exactly as the problem is given. – Nima Dec 10 '12 at 19:22

Something looks odd about this problem.

Lemma If three rows are in arithmetic progression, then they are linearly dependent.

Proof: Let $$r_1=(a_1,a_1+d,..., a_1+(n-1)d) \,;\, r_2=(a_2,a_2+e,..., a_2+(n-1)e) \,;\, r_3=(a_3,a_3+f,..., a_3+(n-1)f)$$


$$er_1=(ea_1,ea_1+ed,..., ea_1+(n-1)ed) \,;\, dr_2=(da_2,da_2+de,..., da_2+d(n-1)e)$$


$$er_1-dr_2=(ea_1-da_2) (1,1,1,...,1)$$

Similarly $$fr_1-dr_3=(fa_1-da_3)(1,1,...,1) \,.$$

From here you can easily get a linear combination which is zero. if any of the coefficients is non-zero you are done, and if all the coefficients are zero you can easily prove directly that these three rows are linearly dependent (since the rows are similar but simpler).

Consequence Let $A$ be any matrix so that the first three rows are anything, and the remaining ones form arithmetic progressions. Then $r(A) \leq 5$

Proof: Assume by contradiction that $R(A) \geq 6$. then we can find 6 linearly independent rows, which then contain three rows which form an arithmetic progression. Contradiction

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