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I have the matrix $G=(I_{16},A)$ where $A=\left( \begin{array}_ J&I_4&I_4&I_4 \\ I_4&J&I_4&I_4 \\ I_4&I_4&J&I_4 \\ I_4&I_4&I_4&J \end{array} \right)$ and $J=\left( \begin{array}_ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1 \end{array} \right)$.

It's difficult at least for me to find the rank of this matrix and more difficult to find the minimum number of the dependent vectors.

Do you have any thoughts of how to do it? Any quick tricks?

Thank you!

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  • $\begingroup$ source of the problem? what techniques have you been studying that led up to this? $\endgroup$
    – Will Jagy
    Aug 3, 2014 at 18:58
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    $\begingroup$ fine. suggest you do the problem in detail for each block $m$ by $m$ for, say, $m \leq 4,$ and $k$ by $k$ blocks, so the whole thing is $mk$ by $mk,$ for $mk$ not as big as 16. maybe some patterns will emerge. $\endgroup$
    – Will Jagy
    Aug 3, 2014 at 19:26
  • $\begingroup$ @WillJagy, this matrix is a generator matrix of a code. I don't have the code in detail though $\endgroup$
    – user113576
    Aug 3, 2014 at 20:11
  • $\begingroup$ Well, if it is important, you will do what I said, examine easier cases of the same general pattern. $\endgroup$
    – Will Jagy
    Aug 3, 2014 at 22:51

1 Answer 1

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Since you need a quick trick... First and trivial note is that $A=A^T$. Second, it is a very simple matrix, ones and zeros. So for 2 columns to be dependent, the need to be identical. I can't see any identical columns, so i would say its a full rank matrix

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  • $\begingroup$ correct about the $A=A^t$ but you are wrong about dependency. Depedency is not only between two vectors $\endgroup$
    – user113576
    Aug 3, 2014 at 17:09
  • $\begingroup$ You are right about it, my bad. I just noted that we need to find rang(G) and not rank(A). In G there are dependencies. (take column(18) = column(17) - column(5) + column(6) ). So G is definatelly not full rank... $\endgroup$ Aug 3, 2014 at 17:19

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