I am working with a linear equation of the following form. x = q + P x ,where x,q are vectors of size K and P is a square matrix of size K. The variables satisfy: (1) q &gt; 0, (2) 0 ≤ P i , j ≤ 1 for all i , j and (3) ∑ j P i , j &lt; 1 for each i.If ( I − P ) is nonsingular, then the equation can be solved by x = ( I − P ) − 1 q. My question is whether or not ( I − P ) is always nonsingular under the above conditions, and if not, what is the condition to add on to make sure the invertibility. I conjecture that it has something to do with the eigenvalues of the matrix but cannot figure out an answer.

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I am working with a linear equation of the following form.  x  =  q  +  P  x  ,where x,q are vectors of size   K and   P is a square matrix of size   K. The variables satisfy: (1)   q  &amp;gt;  0, (2)   0  ≤      P          i      ,      j        ≤  1 for all   i  ,  j and (3)       ∑    j        P          i      ,      j        &amp;lt;  1 for each   i.If   (  I  −  P  ) is nonsingular, then the equation can be solved by   x  =  (  I  −  P      )          −      1        q. My question is whether or not   (  I  −  P  ) is always nonsingular under the above conditions, and if not, what is the condition to add on to make sure the invertibility. I conjecture that it has something to do with the eigenvalues of the matrix but cannot figure out an answer.

Braylon Perez · Accepted Answer

Given those conditions,   (  I  −  P  ) is nonsingular. It is slightly easier to show that   (  I  −  P      )    T    =  I  −      P    T   is nonsingular, and this is sufficient, since then we deduce that  det  (  I  −  P  )  =  det  (  (  I  −  P      )    T    )  =  det  (  (  I  −      P    T    )  ≠  0.Suppose there exists   y  ≠  0 in the kernel of   I  −      P    T  , i.e.   (  I  −      P    T    )  y  =  0. Then   y  =      P    T    y. On             R        K  , let us use the       ℓ    1   norm   ‖  .      ‖    1    :  z  ↦      ∑    1    K        |        z    i        |  . Then  ‖  y      ‖    1    =  ‖      P    T    y      ‖    1    =      ∑          i      =      1        K        |          ∑              j        =        1            K              P              j        ,        i                    y      j        |    .By the triangle inequality,  ‖  y      ‖    1    ≤      ∑          j      =      1        K        |        y    j        |        ∑          i      =      1        K        |        P          j      ,      i            |    &amp;lt;      ∑          j      =      1        K        |        y    j        |    =  ‖  y      ‖    1    .This contradicts   y  ≠  0, so we conclude that   ker  ⁡  (  I  −      P    T    )  =  {  0  } is trivial, and   I  −      P    T   is nonsingular.

Arraryeldergox2 · Accepted Answer

For the system of equations to have a unique solution then the determinant of the matrix   (  I  −  M  ) must be non-zero.Now if it is zero then the system of equations either have infinite solutions or no solution. To find which one, do Gaussian Elimination. If you find that a certain variable can be a free variable ( e.g. 0=0 in augmented matrix) then it has infinite solutions. But, if you find an inconsistency in a variable like 0=-3(or any non zero number) then the system has no solutions.

I am working with a linear equation of the following form. x = q + P x , w

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