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I tried to find matrices that will satisfy the condition of AC = AD when C is not equal D and A has no elements equal to zero, however I could find any right option since when it comes to system of equations it becomes tough since non of the A entries are 0 and I can't find the right elements of C and D. Please help or describe a functioning algorithm of finding those matrices

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  • $\begingroup$ Hint: try choosing an $A$ that maps every vector to a multiple of $(1,1)^T$. $\endgroup$ – amd Sep 15 '18 at 23:27
  • $\begingroup$ Find matrices $R,S$ such that $R$ has no zeros but $RS=0$. Then find matrices $X,Y$ such that $X-Y=S$. $2\times2$ will do. Thinking about the rank of $R$ will help. $\endgroup$ – Gerry Myerson Sep 16 '18 at 0:51
  • $\begingroup$ What’s most important is to start with a matrix $A$ whose determinant is equal to zero. Then your quest should have a happy ending. $\endgroup$ – Lubin Sep 16 '18 at 1:07
  • $\begingroup$ Thanks @ Gerry Myerson , your method works well. Could you explain in liitle more a detail ho did you come to this formulas such as RS = 0 and X-Y = S? Based on what? Is it kind of similar to explanation of Ahmed S. Attaalla? $\endgroup$ – ViniLL Sep 16 '18 at 20:17
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Since $A(C-D)=0$, you just need to find an $A$ with nontrivial null space (maybe that’ll be more clear if you think about $C-D$ as a matrix with vectors $v_1,v_2,..,v_n$ as columns).

As an example how one might do that. Consider the $2$ by $2$ matrix,

$$A=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

Obviously the first row minus the second row gives $\mathbf{0}$, so $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ is in the null space. So let $C$ and $D$ be any two matrices such that:

$$C-D=\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Or,

$$C-D=\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$

Or,

$$C-D=\begin{pmatrix} 1 & 1 & 1 \\ -1 & -1 &-1 \end{pmatrix}$$

Etc...

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  • $\begingroup$ Hello, Ahmed, thanks for you answer, I tried to find C and D matrices based on that and it worked well. I am not sure though if I get right the thing about nontrivial null space , so is it considered to be nontrivial if the difference between row and column is equal to 0 for matrix A? Is that a general rule? And then we just find a vector oh matrice, where row is not equal to coulmn? Let's say, could have we taken C - D = ( 1 -2) instead? $\endgroup$ – ViniLL Sep 16 '18 at 20:10
  • $\begingroup$ Hi, Dan. If you consider the system $Ax=0$, where $x$ is a column vector, the different solutions together form what we call the “Null space of Matrix A”. Notice that $x=0$ will always be a solution, and so it will always be in the null space. We call $x=0$ the “trivial” solution. If we can find an $x$ that is not zero, than we have found a “nontrivial” solution. So what we mean when we say $A$ has nontrivial null-space, is that $x=0$ isn’t the only solution to $Ax=0$. @DanF. Let me know if you need further clarification. Thanks! $\endgroup$ – Ahmed S. Attaalla Sep 16 '18 at 20:15
  • $\begingroup$ So yeah, I think I get it know, in our case we have C-D instead of x in the system Ax = 0? and we have to make sure that C-D is not zero, i.e. non-trivial? $\endgroup$ – ViniLL Sep 16 '18 at 20:20
  • $\begingroup$ That’s right since we need that $C \neq D$ we need to make sure that $C-D \neq 0$, but yet solves $A(C-D)=0$. @DanF. $\endgroup$ – Ahmed S. Attaalla Sep 16 '18 at 20:21
  • $\begingroup$ Got it. Thanks @Ahmed S. Attaalla $\endgroup$ – ViniLL Sep 16 '18 at 20:49

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