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Thread: 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 $\displaystyle n\in\mathbb{N}$, 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 $\displaystyle 1 \neq 0$ 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 $\displaystyle n=1$... but for $\displaystyle n \neq 1$, here goes : it's trivial for $\displaystyle n=0$, and for $\displaystyle n>1$, the determinant $\displaystyle \mod n$ is equal to $\displaystyle 1$, 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 $\displaystyle 1 \neq 0$ 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 $\displaystyle n$ is an even positive integer, then the $\displaystyle n\times n$ matrix $\displaystyle A$ given by $\displaystyle A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$, is invertible.
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  7. #7
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    The inverse is :

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

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

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


    Consider the system of the linear equations :

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

    Let $\displaystyle S $ be the quantity $\displaystyle \sum x_i $ then we have

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

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

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

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

    So $\displaystyle 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 $\displaystyle det(B_m) = (1-n)^{m-1}(nm-n+1)$
    Last edited by simplependulum; Jun 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 $\displaystyle n$ is an even positive integer, then the $\displaystyle n\times n$ matrix $\displaystyle A$ given by $\displaystyle A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$, is invertible.
    Can't $\displaystyle n $ be an odd number $\displaystyle > 1 $ ??

    The inverse is $\displaystyle \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 $\displaystyle n $ be an odd number $\displaystyle > 1 $ ??

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

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

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

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

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

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