# Thread: Non linear map - determinant

1. ## 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:

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

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

2. 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

3. Hello,

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

for example,

$\displaystyle \begin{vmatrix} 1&1&1 \\ 2&2&2 \\ a&a^2&a^3 \end{vmatrix}=a \begin{vmatrix} 1&1&1 \\ 2&2&2 \\ 1&a&a^2 \end{vmatrix}$

4. Thanks for help.

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

greetings

5. 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

6. okay, thanks. I will try it.

greetings

7. Hi,

I stumble again. I subtract bouth columns and get:

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

This looks quite evil but subtraction of columns is an elementary column transforming. Whats next?

8. 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

9. 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?

10. Originally Posted by Herbststurm
Hi,

I stumble again. I subtract bouth columns and get:

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

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
$\displaystyle \begin{bmatrix}0 & 0 1 a-b & b-c & c \\ a^2- b^2 & b^2-c^2 & c^2\end{bmatrix}$

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".

11. 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)

12. okay, but I still don't know how to pull out the c-square...

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

I think the problem is that you want laplace development and I am not allowed to do this.

13. Originally Posted by Herbststurm
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:

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

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 $\displaystyle \boxed{w(Ax_1,Ax_2,Ax_3) = \det(A) w(x_1,x_2,x_3)}$, where w is any alternating k-multilinear map. Note that the map is well defined(Why?)

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

Now using the fact that w is linear, we can break $\displaystyle w\left(\sum_{k=1}^{k=3} a^{k-1} x_k, \sum_{k=1}^{k=3} b^{k-1} x_k,\sum_{k=1}^{k=3} c^{k-1} x_k\right)$ 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
$\displaystyle 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)$$\displaystyle = w(x_1,bx_2,c^2x_3) + w(ax_2,x_1,c^2x_3) + w(a^2x_3,bx_2,x_1)$ $\displaystyle + w(x_1,b^2x_3,cx_2) + w(ax_2,b^2x_3,x_1) + w(a^2 x_3,x_1,cx_2)$

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

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

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

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

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

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

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

But by definition, $\displaystyle w(Ax_1,Ax_2,Ax_3) = \det(A) w(x_1,x_2,x_3)$ and thus $\displaystyle \boxed{\det (A) = (a-b)(b-c)(c-a)} \,\,\,\,\,\,\,\,\, \blacksquare$

14. That was what I need.