Consider the (real) linear system of equations $A\mathbf{x}=\mathbf{c}$ of size $N$ as $$ \begin{bmatrix} a_{N}-a_{2}& a_{2}& 0 &\dots& 0 & -a_{N} \\-a_{1} & a_{1}-a_{3}&a_{3}& 0 & \dots&0\\ 0 & -a_{2} & a_{2}-a_{4}&a_{4}&0 & \dots \\\vdots&\ddots&\ddots&\ddots&\ddots\\ a_1& 0& \dots& 0&-a_{N-1}&a_{N-1}-a_1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\\\vdots\\x_N\end{bmatrix}=\begin{bmatrix} c\\c\\c\\\vdots\\c\end{bmatrix}. $$ So, the vector $\mathbf{c}$ is constant and $A$ has a periodic structure and its kernel (at least) contains constant vectors (since the sum of columns is zero).

How can one prove that

  1. $rank(A)=N-1$
  2. The system is inconsistent, unless $c=0$.

I suppose that these claims are valid. But any counter-example is also very appreciated.

  • 1
    $\begingroup$ If $a_i=0$ for all $i$, then the rank of $A$ is zero. Are there any extra conditions for $a_i$? $\endgroup$ – i707107 Dec 8 '16 at 19:28
  • $\begingroup$ @i707107 Yes, zero matrix is so trivial. I can additionally assume that all $a_i$ are either positive or negative. $\endgroup$ – Hamed Dec 9 '16 at 8:40
  • $\begingroup$ In that case, the rank is either $N-2$ or $N-1$, but I couldn't disprove the case $N-2$. Maybe there is a way.. $\endgroup$ – i707107 Dec 10 '16 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.