Non linear map - determinant

• Jan 15th 2009, 12:45 PM
Herbststurm
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. (Wondering)

thanks
greetings
• Jan 15th 2009, 01:15 PM
ursa
Quote:

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:

http://www.mathhelpforum.com/math-he...f26d73f9-1.gif

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. (Wondering)

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

this will make it easier(Wink)
• Jan 15th 2009, 01:29 PM
Moo
Hello,

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

for example,

$\begin{vmatrix} 1&1&1 \\ 2&2&2 \\ a&a^2&a^3 \end{vmatrix}=a \begin{vmatrix} 1&1&1 \\ 2&2&2 \\ 1&a&a^2 \end{vmatrix}$
• Jan 15th 2009, 02:01 PM
Herbststurm
Thanks for help.

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

greetings
• Jan 15th 2009, 10:46 PM
ursa
Quote:

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
• Jan 15th 2009, 10:49 PM
Herbststurm
okay, thanks. I will try it.

greetings
• Jan 16th 2009, 02:33 PM
Herbststurm
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? (Thinking)
• Jan 16th 2009, 08:01 PM
ursa
Quote:

Hi,

I stumble again. I subtract bouth columns and get:

http://www.mathhelpforum.com/math-he...e3c20a8a-1.gif

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

Quote:

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
• Jan 17th 2009, 01:13 AM
Herbststurm
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? (Thinking)
• Jan 17th 2009, 03:20 AM
HallsofIvy
Quote:

Originally Posted by Herbststurm
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? (Thinking)

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".
• Jan 17th 2009, 03:55 AM
ursa
Quote:

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? (Thinking)
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)
• Jan 17th 2009, 03:58 AM
Herbststurm
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.
• Jan 17th 2009, 04:02 AM
Isomorphism
Quote:

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:

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

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. (Wondering)

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
$w\left(\sum_{k=1}^{k=3} a^{k-1} x_k, \sum_{k=1}^{k=3} b^{k-1} x_k,\sum_{k=1}^{k=3} c^{k-1} x_k\right)$ $
= 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 $\boxed{\det (A) = (a-b)(b-c)(c-a)} \,\,\,\,\,\,\,\,\, \blacksquare$
• Jan 17th 2009, 04:26 AM
Herbststurm