Linear Algebra geniuses come in...

blustori

Senior member
Mar 2, 2005
753
0
0
How can I prove that the det of an nxn matrix != 0 when each element is defined by w^nk and w = e^-2*pi*j/N where N = the length of the matrix? Hint: the first row and column are all 1's. (I was going to write a program to do this, but then I wouldn't be able to see the pattern.) This is the kind of problem I wish I could solve in a snap.

I am staying up until I get this... with(out) the help of ATOT! I really need this extra credit. TIA
 

KLin

Lifer
Feb 29, 2000
30,500
787
126
Do your own @#$@#$@ homework you lazy @#$(@#( son of a @!$(@#$(@ !@#!^&.
 

RGUN

Golden Member
Dec 11, 2005
1,007
3
76
why dont you write out a 3x3 and a 4x4 and go from there? they wont take that long to solve and you should be able to recognize any patterns easily enough. I would try and help more but I havent taken linear Algebra in two years.
 

OdiN

Banned
Mar 1, 2000
16,430
3
0
What if the det of an nxn matrix really does = 0 and it's a trick question?


EDIT:

BTW I have no clue what the abbreviations det or nxn mean.

I was lucky to pass pre-calc. Math gets really damn confusing when you're dyslexic.
 

blustori

Senior member
Mar 2, 2005
753
0
0
k is the same as n n,k == row, column. The pattern I see is for example
[ 1 1 1 1 ]
[ 1 a b c ]
[ 1 b d e ]
[ 1 c e f ]
hard to explain... I need to look at it more.
 

Howard

Lifer
Oct 14, 1999
47,982
11
81
If you want the prog to calculate the det of a given 3x3 or 4x4 matrix, hit me up tomorrow morning.
 

Hyperion042

Member
Mar 23, 2003
53
0
0
It's clearly symmetric, therefore it's diagonalizable, therefore its determinant is equal to the product of its eigenvalues, all of which should be positive in that.

...Maybe. That's what I get at first glance.
 

kogase

Diamond Member
Sep 8, 2004
5,213
0
0
Originally posted by: Hyperion042
It's clearly symmetric, therefore it's diagonalizable, therefore its determinant is equal to the sum of its eigenvalues, all of which should be positive in that.

...Maybe. That's what I get at first glance.

Yes, I concur.
 

alphatarget1

Diamond Member
Dec 9, 2001
5,710
0
76
Originally posted by: blustori
How can I prove that the det of an nxn matrix != 0 when each element is defined by w^nk and w = e^-2*pi*j/N where N = the length of the matrix? Hint: the first row and column are all 1's. (I was going to write a program to do this, but then I wouldn't be able to see the pattern.) This is the kind of problem I wish I could solve in a snap.

I am staying up until I get this... with(out) the help of ATOT! I really need this extra credit. TIA

w^nk where n = the length/width of the matrix? (square, same number)

gotta be more specific... i don't foresee myself solving this problem even though i took that class already. got a C in it.