# Math Help - Linear Algebra

1. ## Linear Algebra

Help.. i can't understand the meaning... May i know how to form the matrix R? Thanks in advance

The Nth root of 1 is given by w=exp(2*pi*j / N), where j = sqrt -1. It is easily verified as w^N = exp (2*pi*j) = cos 2pi + j sin2pi = 1. Using this N th root of 1, we create an (N*N) square matrix R such that the (a,b) th element of R is given by rab = W^(a-1)(b-1), a = 1,2,....,N; b = 1, 2, ...., N.

Question--> Write the matrix R in terms of it's element. Clearly, show at least the top 3 X 3 part and all element on the four corner...

Is this symmetric??

2. Originally Posted by Chris0724
Is this symmetric??
You need to show $a_{ij} = a_{ji}$.
Note, $a_{ij} = \omega^{(i-1)(j-1)} = \omega^{(j-1)(i-1)} = a_{ji}$.
Thus it is symmetric.

3. i had form the matrix.

| 1 1 1..........1|
| 1 w^1 w^2 |
| 1 w^2 w^4 |
| .
| .
|1............W^(n-1)(n-1)|

Is the matrix above correct if i leave the N-th row & N-th coloum like that?

Qns 2
Consider another (N × N) square matrix S such that the (c, d)-th element of S is given by Scd = w^(c-1)(d-1), c = 1,2,....,N; d = 1, 2, ...., NShow that determinant of S is equal to the complex conjugate of the determinant of R. Note all you need to show is that S and R are complex conjugate of each other. In other words, show that ω and ω1 are complex conjugate of each other.

Is this correct?

w^N = exp (2*pi*j) = cos 2pi + j sin2pi = 1
w^N = exp (2*pi*j) = cos 2pi - j sin2pi = -1

Qns 3
Let T = SR. Write the elements of T.

May i know wat is w^-2?

| 3 1+w^-1+w^-2 ........
|

Once again many thank you for the help

hi,