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Math Help - Non linear map - determinant

  1. #1
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    Non linear map - determinant

    Hi,

    How to calculate a determinant if the coefficients are not linear? This confuses me because the determinant map is an alternating k-multilinear map.

    Something like:

    det(\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix})

    in the third row are the coefficients quadratic and not linear. I guess I am not allowed to use the rules I know, for example Sarrus rule.

    thanks
    greetings
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  2. #2
    Junior Member ursa's Avatar
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    Hi,

    How to calculate a determinant if the coefficients are not linear? This confuses me because the determinant map is an alternating k-multilinear map.

    Something like:



    in the third row are the coefficients quadratic and not linear. I guess I am not allowed to use the rules I know, for example Sarrus rule.

    thanks
    greetings
    hi
    do the followng steps:
    column1-column2
    column2-column3

    this will make it easier
    Last edited by ursa; January 15th 2009 at 01:36 PM.
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  3. #3
    Moo
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    Hello,

    Following what ursa said, if you have a common factor on a row, you can factor it out of the determinant :

    for example,

    \begin{vmatrix} 1&1&1 \\ 2&2&2 \\ a&a^2&a^3 \end{vmatrix}=a \begin{vmatrix} 1&1&1 \\ 2&2&2 \\ 1&a&a^2 \end{vmatrix}
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  4. #4
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    Thanks for help.

    @ursa
    what do you mean with your hint. Interchange of the columns?

    greetings
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  5. #5
    Junior Member ursa's Avatar
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    Thanks for help.

    @ursa
    what do you mean with your hint. Interchange of the columns?

    greetings
    hi
    sorry for the ambiguity
    bt '-' is not for interchange, it is for subtraction
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  6. #6
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    okay, thanks. I will try it.

    greetings
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  7. #7
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    Hi,

    I stumble again. I subtract bouth columns and get:

    det(\begin{pmatrix} 1 & 0 & -1 \\ a & a-b & a-b-c \\ a^2 & a^2-b^2 & a^2-b^2-c^2 \end{pmatrix})

    This looks quite evil but subtraction of columns is an elementary column transforming. Whats next?
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  8. #8
    Junior Member ursa's Avatar
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    Hi,

    I stumble again. I subtract bouth columns and get:



    This looks quite evil but subtraction of columns is an elementary column transforming. Whats next?
    hi
    i m sorry for writing only....

    do the followng steps:
    column1-column2
    column2-column3
    actually it is
    column1->column1-column2
    column2->column2-column3
    so marix comes out as
    ( 0 0 1 )
    ( a-b b-c c )
    ( (a^2)-(b^2) (b^2)-(c^2) c^2)

    now in third row you can apply formula of (a^2)-(b^2)
    and then you can take out common from the column

    sorry,but i dnt knw the coding
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  9. #9
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    But what happend with the thierd column?

    The first and second coloum is clear but the third has in the first row an one. I can not pull out the c-square because I wouldn't get the one. How to fix this problem?
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  10. #10
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    Quote Originally Posted by Herbststurm View Post
    Hi,

    I stumble again. I subtract bouth columns and get:

    det(\begin{pmatrix} 1 & 0 & -1 \\ a & a-b & a-b-c \\ a^2 & a^2-b^2 & a^2-b^2-c^2 \end{pmatrix})

    This looks quite evil but subtraction of columns is an elementary column transforming. Whats next?
    I don't believe that was what was meant by "column1-column2
    column2-column3"
    If you subtract the second column from the first and subtract the third column from the original second column, you get
    \begin{bmatrix}0 & 0 1  a-b & b-c & c \\ a^2- b^2 & b^2-c^2 & c^2\end{bmatrix}

    It's not too difficult then to see that the determinant is (a-b)(b-c)(c-a).

    I don't know why you said you were "not allowed to use the rules I know".
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  11. #11
    Junior Member ursa's Avatar
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    But what happend with the thierd column?

    The first and second coloum is clear but the third has in the first row an one. I can not pull out the c-square because I wouldn't get the one. How to fix this problem?
    hi
    you dont have to pull out anything from third column
    after taking common from first and second column
    calculate determinant along first row
    since first two values are 0
    therefore you jst have to calculate with the value one (first row, third column)
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  12. #12
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    okay, but I still don't know how to pull out the c-square...

    det(\begin{pmatrix} 0 & 0 & 1 \\ a-b & b-c & c \\ a^2-b^2 & a^2-c^2 & c^2 \end{pmatrix})

    I think the problem is that you want laplace development and I am not allowed to do this.
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  13. #13
    Lord of certain Rings
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    Quote Originally Posted by Herbststurm View Post
    Hi,

    How to calculate a determinant if the coefficients are not linear? This confuses me because the determinant map is an alternating k-multilinear map.

    Something like:

    det(\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix})

    in the third row are the coefficients quadratic and not linear. I guess I am not allowed to use the rules I know, for example Sarrus rule.

    thanks
    greetings
    I am writing this long proof because you said k-multilinear map and that you are not allowed to use Sarrus rule. This way is long winded compared to what others have said, but I think this what is expected from you:

    We know that the determinant of a linear transformation A is defined as \boxed{w(Ax_1,Ax_2,Ax_3) = \det(A) w(x_1,x_2,x_3)}, where w is any alternating k-multilinear map. Note that the map is well defined(Why?)

    Now choose x_1, x_2, x_3 to be the standard basis. Then w(Ax_1,Ax_2,Ax_3) = w\left(\sum_{k=1}^{k=3} a^{k-1} x_k, \sum_{k=1}^{k=3} b^{k-1} x_k,\sum_{k=1}^{k=3} c^{k-1} x_k\right)

    Now using the fact that w is linear, we can break w\left(\sum_{k=1}^{k=3} a^{k-1} x_k, \sum_{k=1}^{k=3} b^{k-1} x_k,\sum_{k=1}^{k=3} c^{k-1} x_k\right) as the sum of 27 w(.,.,.) terms. But since w() is alternating, only 6 terms will survive, the ones in which all the arguments are distinct. Thus
    w\left(\sum_{k=1}^{k=3} a^{k-1} x_k, \sum_{k=1}^{k=3} b^{k-1} x_k,\sum_{k=1}^{k=3} c^{k-1} x_k\right) <br />
= w(x_1,bx_2,c^2x_3) + w(ax_2,x_1,c^2x_3) + w(a^2x_3,bx_2,x_1) + w(x_1,b^2x_3,cx_2) + w(ax_2,b^2x_3,x_1) + w(a^2 x_3,x_1,cx_2)

    But since w() is linear, we can pull the constants out:
    w(Ax_1,Ax_2,Ax_3) = bc^2w(x_1,x_2,x_3) + ac^2w(x_2,x_1,x_3) + a^2bw(x_3,x_2,x_1) + b^2c w(x_1,x_3,x_2) + ab^2 w(x_2,x_3,x_1) + a^2c w(2 x_3,x_1,x_2)

    Now use the fact that an alternating k-multilinear map is skew symmetric. Thus if \pi \in S_3 the symmetric group of 3 letters, w(x_{\pi(1)},x_{\pi(2)},x_{\pi(3)}) = \text{sign}(\pi)w(x_1,x_2,x_3)

    So we get:
    w(x_1,x_3,x_2) = w(x_2,x_1,x_3) = w(x_3,x_2,x_1)= - w(x_1,x_2,x_3)
    w(x_2,x_3,x_1) = w(x_3,x_1,x_2) = w(x_1,x_2,x_3)

    Thus w(Ax_1,Ax_2,Ax_3) = bc^2w(x_1,x_2,x_3) - ac^2w(x_1,x_2,x_3) - a^2bw(x_1,x_2,x_3) - b^2c w(x_1,x_2,x_3) + ab^2 w(x_1,x_2,x_3) + a^2c w(x_1,x_2,x_3)

    Thus w(Ax_1,Ax_2,Ax_3) = (bc^2 - ac^2 - a^2b - b^2c + ab^2 + a^2c) w(x_1,x_2,x_3)

    But (bc^2 - ac^2 - a^2b - b^2c + ab^2 + a^2c)  = (a-b)(b-c)(c-a)

    Thus w(Ax_1,Ax_2,Ax_3) = (a-b)(b-c)(c-a) w(x_1,x_2,x_3)

    But by definition, w(Ax_1,Ax_2,Ax_3) = \det(A) w(x_1,x_2,x_3) and thus \boxed{\det (A) = (a-b)(b-c)(c-a)} \,\,\,\,\,\,\,\,\, \blacksquare
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  14. #14
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    That was what I need.

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