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A system of equations with more unknowns than equations

  1. Sep 13, 2010 #1

    It is known that a system of linear equations is considered overdetermined if there are more equations than unknowns...

    My question simply stated: is there a term to describe a system of linear equations if there are more equations than unknowns?
    Google seems to struggle when my search query is
    "A system of equations with more unknowns than equations" !

    Ive recently encountered such a system and was curious to do more research into developing or looking into efficient algorithms for solving them---since my current method of solving them is a bit tedious.

    Thanks for any information.
  2. jcsd
  3. Sep 13, 2010 #2


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  4. Sep 13, 2010 #3
    hmm...Ive gotten some hits based off of using that terminology (interesting how it isnt mentioned in relation to an overdetermined sytem)

    Some initial reading indicates:
    Note that an underdetermined system might be either consistent or inconsistent, depending on the equations

    You happen to know what consistent vs inconsistent means?
  5. Sep 13, 2010 #4


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    "Consistent" means that there exist at least one solution. "Inconsistent" means there is no solution.

    Examples with 3 unknowns and 2 equations are:
    x+ y+ z= 2, x+ y- z= 4: consistent

    and x+ y+ z= 2, x+ y+ z= 1: inconsistent

    With x+ y+ z= 2 and x+ y- z= 4, I can add the equations to eliminate z and get 2x+ 2y= 6 so that x+ y= 3. I can choose any value for, say, x, then solve for y, then z. For example, if x= 1, then y= 3- 1= 2 so that the first equation becomes 1+ 2+ z= 3+ z= 2 and z= -1. x= 1, y= 2, z= -1 is a solution. But I could as well, take x= 2 and get y= 3- 2= 1, 3- z= 4, x= -1. In fact, (x, 3- x, -1) is a solution for x any real number.

    With x+ y+ z= 2, x+ y+ z= 1, subtracting the second equation from the first, we get 0= 1 which is impossible. If fact the two left sides are identical. If x+ y+ z= 2, then it can't possibly be equal to 1 whatever x and y are!
  6. Sep 13, 2010 #5
    I have come across:
    deficient for not enough equations
    sufficient for just enough, and
    redundant for too many

    Is that helpful?
  7. Sep 13, 2010 #6
    Thank you very much for the terminology clarification, thats what I thought the terms meant---seems different sources use different ways to describe the same thing at times.

    I appreciate the help....I would like to post my method here when I establish something to work these 'underdetermined' systems (not a proof mind you!) just to make sure Im not reinventing the wheel or anything (in addition to taking unnecessary steps)

    thanks again for the help with all my questions
  8. Sep 14, 2010 #7
    What you're attempting has probably already been shown in mxn matrices where m<n.
  9. Sep 14, 2010 #8


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    A common technique of solving underdetermined systems is to find the least norm solution (i.e. the solution with the smallest euclidean norm). Perform a Google search on 'least norm solution'.
  10. Sep 14, 2010 #9
    Thanks! I had not heard that term before although Least Squares has come up before and I have looked into that algorithm before---have not read through it yet but I am guessing thaey must be similar.

    One method I was shown in n x n matices is to have your
    variable matrix

    and a

    constant matrix
    then find determinant of the variable matrix, and let it be 'k'

    Next to get value of x replace 1st column with constant matrix -> and find the determinant, and let it be 'c', therefore the value of x is 'c/k' so on and so forth for y and z.

    My question then is this method applicable to an n x m matrix? I have not taken linear algebra yet, but I do know than inverses do not exist for rectangular matrices (is this why pseduinverses are used???) but do determinants?

    Thanks for the input...this is highly engaging and I enjoy coming across new tools
  11. Sep 15, 2010 #10
    The method you describe, which is known as Cramer's rule, follows from the adjugate formula for the inverse of a matrix; that is, A-1 = adj(A)/det(A). (You can prove Cramer's rule from this by expanding the top determinant along the column you replaced.) However, adjugates and determinants only exist for square matrices.

    I don't know of any similar method for non-square matrices that is equivalent to using the pseudoinverse. In particular, the pseudoinverse depends on the inner product (since, after all, for an equation Ax = b the pseudoinverse gives the vector x = A+b which minimizes the norm of x or Ax - b, depending on whether the system is underdetermined or overdetermined, but x depends on the norm), so any such method must use the inner product somewhere.
  12. Sep 15, 2010 #11
    Thanks for giving a name to that method.

    Consider then this system of 2 linear equations:

    ( (-2*sqrt(3) )/9 )x + z = 2*sqrt(3)

    3*sqrt(2)y - 2x = 0

    Here are presented two equations with 3 unknowns (x,y,z) is this system solvable using a matrix method? The reason being is because none of the equations have ALL 3 variables in them only two at a time. Is it a necessary condition that requires all three variables to be in the linear equations to be solved using a matrix algorithm?
    Therefore does this constitute by def. a linear system of equations that is solvable using a known matrix method as we have been discussing above.
    All example Ive come across seem to have all variables present in each linear equation of the system.

    Thanks for the discussion
  13. Sep 15, 2010 #12
    Of course. Just set 0s where the "missing" variables are.
  14. Sep 16, 2010 #13


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    This is equivalent to the matrix equation
    [tex]\begin{bmatrix}-\frac{2\sqrt{3}}{9} & 0 & 1 \\ -2 & 3\sqrt{2} & 0\end{bmatrix}= \begin{bmatrix}2\sqrt{3}\\ 0\end{bmatrix}[/tex]
  15. Feb 27, 2012 #14
    In this particular case, if we use the Ax=b method, where we then invert A (say with A-1 and premultiply b with this, then we must have that A has an inverse. But A here is not a square matrix, and only square matrices have inverses, so the matrix solutions will not prove fruitful.

    On the other hand, for all my spoofery, I am trying to prove that the only solution to α1 -2α2 +3α3=0 and 2α23=0 is that α123=0. Can you offer any help?
  16. Feb 27, 2012 #15


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    Staff: Mentor

    This is not the only solution.
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