For each m , n ∈ N think R m as the space of real column vectors of size m and R m × n as the space of matrices of size m × n.Let d ∈ NLet a : { 1 , … , 2 d } → { 1 } × { 0 , 1 } d be an enumeration (injective and surjective map).Let A ∈ R ( d + 1 ) × 2 d be the matrix whose columns are a ( 1 ) , … , a ( 2 d )Is it true that for each b ∈ { 1 } × [ 0 , 1 ] d there exists x ∈ [ 0 , + ∞ ) 2 d such that A x = b ??

Question

For each   m  ,  n  ∈  N think         R    m   as the space of real column vectors of size   m and         R          m      ×      n       as the space of matrices of size   m  ×  n.Let   d  ∈  NLet   a  :  {  1  ,  …  ,      2    d    }  →  {  1  }  ×  {  0  ,  1      }    d   be an enumeration (injective and surjective map).Let   A  ∈        R          (      d      +      1      )      ×              2        d             be the matrix whose columns are   a  (  1  )  ,  …  ,  a  (      2    d    )Is it true that for each   b  ∈  {  1  }  ×  [  0  ,  1      ]    d   there exists   x  ∈  [  0  ,  +  ∞      )            2      d       such that   A  x  =  b  ??

Carmelo Payne · Accepted Answer

Yes, there is an   x with       x    i    ∈  [  0  ,  1  ].Essentially, you want to solve      ∑          n      =      1                2      d            x    n    =  1  ,        ∑          ⌊              n                  2          k                    ⌋      ≡      1            (      mod            2      )            x    n    =      b    k  and you can see      x    n    =      ∏          k      =      1        d        f    k    (  n  )  ,        f    k    (  n  )  :=      {                                        b            k                                    ⌊                      n                          2              k                                ⌋          ≡          1                    (          mod                    2          )                                      1          −                      b            k                                    ⌊                      n                          2              k                                ⌋          ≡          0                    (          mod                    2          )                        is a solution.

For each m , n ∈ <mi mathvariant="double-struck">N think

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