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If Ax = 0 has one solution, show m>=n

  1. Jun 8, 2005 #1

    WY

    User Avatar

    Hey

    I'm not to sure on how to approach this problem - I think I have to use induction but I don't know where to start!
    Question: Let A be an mxn matric. Suppose that Ax=0 has only one solution, namely x=0.
    Show that m >= n


    Thanks for the help in advance
     
  2. jcsd
  3. Jun 8, 2005 #2
    Dimension theorem for Matrices says if matrix has n columns => rank(A)+nullity(A)=n
    In your case nullity(A)=dimKer(A)=0
    Is it possible rank(A) to be > min(m,n)???
     
  4. Jun 8, 2005 #3

    WY

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    Thanks for replying :)
    But I'm not to sure what you mean - i don't think i am very familiar with that theorem... could you elaborate a little? hehehe
     
  5. Jun 8, 2005 #4

    OlderDan

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    Science Advisor
    Homework Helper

    If you are unfamiliar with those formal properties and want to approach the problem from the basics, you could start by writing simple cases

    [tex]
    \left[ {\begin{array}{*{20}c}
    a & b \\
    \end{array}} \right]\left[ {\begin{array}{*{20}c}
    {x_1 } \\
    {x_2 } \\
    \end{array}} \right] = 0
    [/tex]


    [tex]
    \left[ {\begin{array}{*{20}c}
    {\begin{array}{*{20}c}
    a & b & c \\
    \end{array}} \\
    {\begin{array}{*{20}c}
    d & e & f \\
    \end{array}} \\
    \end{array}} \right]\left[ {\begin{array}{*{20}c}
    {x_1 } \\
    {x_2 } \\
    {x_3 } \\
    \end{array}} \right] = 0
    [/tex]

    It's pretty easy to show that these equations can be satisfied with non-zero x vectors. From a linear equations perspective, it is matter of having too many unkowns and not enough equations.
     
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