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Math Help - Invertible matrix

  1. #1
    Moo
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    Invertible matrix

    Hi,

    This is probably a simple problem, but I just fell in love with that

    Prove that for any n\in\mathbb{N}, the square matrix consisting in only 1's in the diagonal and n everywhere else is always invertible.
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  2. #2
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    Here is simple proof: The determinant of the matrix will always be 1 \neq 0 hence the matrix will be invertible.
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    MHF Contributor Bruno J.'s Avatar
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    It's not true for n=1... but for n \neq 1, here goes : it's trivial for n=0, and for n>1, the determinant \mod n is equal to 1, so it can't be zero.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by p0oint View Post
    Here is simple proof: The determinant of the matrix will always be 1 \neq 0 hence the matrix will be invertible.
    Beat me by a minute!
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  5. #5
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    Yes, sorry, it's only for n>1.
    And indeed, the simple way to prove it is to look at the matrix mod n
    In *hard* times like these, I like algebra
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Algebra is great!

    Here's a similar problem, perhaps a little harder.

    If n is an even positive integer, then the n\times n matrix A given by A_{ij}=1 \  (i \neq j), \: \: A_{ii}=0, is invertible.
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  7. #7
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    The inverse is :

     -\frac{1}{n-1} I + \frac{n}{(n-1)(nm-n+1) } A

    where  [A]_{ij} = 1 ~~  1 \leq i,j \leq m  and  m is the size of the matrix .

    To obtain the determinant , not only the residue ...


    Consider the system of the linear equations :

      [ nA - (n-1)I ]x = y ~~  , ~  x = (x_1 ,x_2 ,...,x_m )^T ~,~ y = (0,0,0,...,0,1)^T and note that  nA - (n-1)I is your matrix , let it be  B_m ( If i read the problem correctly )

    Let  S be the quantity  \sum x_i then we have

     nS - (n-1) x_i = 0 ~,~ 1 \leq i \leq m-1  ~~,~~ nS - (n-1)x_m = 1

     x_i = \frac{nS}{n-1} ~,~ 1 \leq i \leq m-1 ~~,~~ x_m = \frac{nS - 1}{n-1}

     \frac{n(m-1)S}{n-1} + \frac{nS - 1}{n-1} = S

     S = \frac{1}{nm - n + 1 }

    So  x_m = - \frac{n(m-1)-n+1}{(n-1)(nm-n+1) } = \frac{det(B_{m-1})}{det(B_m)}


    I believe it is true that  det(B_m) = (1-n)^{m-1}(nm-n+1)
    Last edited by simplependulum; June 30th 2010 at 09:56 PM.
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  8. #8
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    Quote Originally Posted by Bruno J. View Post
    Algebra is great!

    Here's a similar problem, perhaps a little harder.

    If n is an even positive integer, then the n\times n matrix A given by A_{ij}=1 \  (i \neq j), \: \: A_{ii}=0, is invertible.
    Can't  n be an odd number  > 1 ??

    The inverse is   \frac{1}{n-1} A - I
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  9. #9
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    Quote Originally Posted by simplependulum View Post
    Can't  n be an odd number  > 1 ??

    The inverse is   \frac{1}{n-1} A - I

    The above doesn't work even for n=2: A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\Longrightar  row A^{-1}=A\neq A-I .

    Spoiler:
    Let's try to find out what conditions must be met so that A\underline{x}=\underline{0}\,,\,\,\underline{x}=\  begin{pmatrix}x_1\\x_2\\\ldots\\x_n\end{pmatrix} :

    \begin{pmatrix}0&1&\ldots&1&1\\1&0&1&\ldots &1\\\ldots&\ldots&\ldots&\ldots&\ldots\\1&1&\ldots  &1&0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\\ldots\\x_n\end{pmatrix}=  \begin{pmatrix}x_2+x_3+\ldots+x_n\\x_1+x_3+\ldots+  x_n\\\ldots\\x_1+x_2+\ldots +x_{n-1}\end{pmatrix}.

    If we equal the above to the zero vector and solve we'll find at once that it must be that x_1=x_2=\ldots =x_n=0 , which means A is 1-1 as linear transformation and thus regular.

    Tonio
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  10. #10
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    Oh , the matrix  A is based on what i defined in the previous post (#7) , sorry .
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