Determinant containing entries of the form cos(a_1 - b_1): Prove that its value is zero.

[Violette]

No no, n can NOT be infinity. It can be as large as you like, but not infinity.

What's the difference? why n cannot be infinity ( according to blamocur n value is bounded between [math]3≤ n < 100[/math] but I don't understand how did he do that...)

 

[blamocur]

what you meant by saying numerical script and symbolic script?

I used SymPy script/program for symbolic computation and NumPy for numerical one. In the latter case I generated matrices for random values of [imath]a[/imath] and [imath]b[/imath], and they all yielded determinant of zero for [imath]n > 2[/imath].

 

[blamocur]

hmn somehow I haven't learnt vector multiply with a matrix but I'll try to do it now I may give you an answer 2moro if you don't mind

But do you know that if for some vector [imath]v[/imath] we have [imath]Mv=0[/imath] or [imath]vM=0[/imath], and [imath]v \neq 0[/imath] then [imath]\det(M) = 0[/imath] ?

 

[Steven G]

What's the difference? why n cannot be infinity ( according to blamocur n value is bounded between [math]3≤ n < 100[/math] but I don't understand how did he do that...)

IF \(\displaystyle 3≤ n < 100\), then how can n be infinity?

You are given an nxn matrix. n is a positive integer. Since oo is NOT a positive integer (oo is not even a number!), then n can't be infinity. A non matrix might be a 2x2 matrix, a 125X125 matrix, 10,123,321x10,123,321 matrix, ... But not ooxoo.

 

[Violette]

Hint: for [imath]n=3[/imath] it is not too difficult to construct a vector [imath]v_0[/imath] such that [imath]v_0 M = 0[/imath].

for n = 3 we have two rows are the same which means Det is equal to 0

 

[topsquark]

View attachment 36651
for n = 3 we have two rows are the same which means Det is equal to 0

For n = 3, the matrix is
[imath]\begin{pmatrix} cos(a_1 - b_1) & cos(a_2 - b_1) & cos(a_3 - b_1) \\ cos(a_1 - b_2) & cos(a_2 - b_2) & cos(a_3 - b_2) \\ cos(a_1 - b_3) & cos(a_2 - b_3) & cos(a_3 - b_3) \end{pmatrix}[/imath]

There will never be duplicate rows.

-Dan

 

[Steven G]

View attachment 36651
for n = 3 we have two rows are the same which means Det is equal to 0

I do have one question, if n=3, then why are there 4 rows and 4 columns?

 

[blamocur]

I do have one question, if n=3, then why are there 4 rows and 4 columns?

Because it is easier to get duplicates ?

 

[Steven G]

Because it is easier to get duplicates ?

Sounds reasonable.

 

[Violette]

For n = 3, the matrix is
[imath]\begin{pmatrix} cos(a_1 - b_1) & cos(a_2 - b_1) & cos(a_3 - b_1) \\ cos(a_1 - b_2) & cos(a_2 - b_2) & cos(a_3 - b_2) \\ cos(a_1 - b_3) & cos(a_2 - b_3) & cos(a_3 - b_3) \end{pmatrix}[/imath]

There will never be duplicate rows.

-Dan

I don't understand. Why is n = 3 the Determinant will have that form?

 

[khansaheb]

I don't understand. Why is n = 3 the Determinant will have that form?

Can you propose a different form for n=3 ?

a_n and b_n stops existing after n=3 (for 0<n<4) - look at your OP carefully.

 

[Violette]

Can you propose a different form for n=3 ?

a_n and b_n stops existing after n=3 (for 0<n<4) - look at your OP carefully.

I see, I misunderstood the 'n'

 

[blamocur]

Claim:
[math]D \left( \begin{array}{l} \sin{(b_2 - b_3)} \\ \sin{(b_3 - b_1)} \\ \sin{(b_1 - b_2)} \end{array} \right) = \left(\begin{array}{c}0 \\ 0 \\ 0\end{array} \right)[/math]where D =
[math]\left( \begin{array}{lll} \cos\left(a_{1}-b_{1}\right) & \cos\left(a_{1}-b_{2}\right) & \cos\left(a_{1}-b_{3}\right) \\ \cos\left(a_{2}-b_{1}\right) & \cos\left(a_{2}-b_{2}\right) & \cos\left(a_{2}-b_{3}\right) \\ \cos\left(a_{3}-b_{1}\right) & \cos\left(a_{3}-b_{2}\right) & \cos\left(a_{3}-b_{3}\right) \end{array} \right)[/math]Note that this proves that [imath]\det D = 0[/imath] only if [imath]v \neq 0[/imath]. But [imath]v=0[/imath] only when [imath]b_1 = b_2 = b_3 \;mod\; 2\pi[/imath], in which case all columns of [imath]D[/imath] are identical, which also makes [imath]\det D = 0[/imath].

Let
[math]\left( \begin{array}{lll} \cos\left(a_{1}-b_{1}\right) & \cos\left(a_{1}-b_{2}\right) & \cos\left(a_{1}-b_{3}\right) \\ \cos\left(a_{2}-b_{1}\right) & \cos\left(a_{2}-b_{2}\right) & \cos\left(a_{2}-b_{3}\right) \\ \cos\left(a_{3}-b_{1}\right) & \cos\left(a_{3}-b_{2}\right) & \cos\left(a_{3}-b_{3}\right) \end{array} \right) \left( \begin{array}{l} \sin{(b_2 - b_3)} \\ \sin{(b_3 - b_1)} \\ \sin{(b_1 - b_2)} \end{array} \right) = \left(\begin{array}{c}x_1 \\ x_2 \\ x_3\end{array} \right)[/math] where
[math]x_i = \cos\left(a_{i}-b_{1}\right) \sin{(b_2-b_3)} + \cos\left(a_{i}-b_{2}\right) \sin{(b_3-b_1)} + \cos\left(a_{i}-b_{3}\right) \sin{(b_1-b_2)}[/math]Using identity [imath]\;\;2\cos u \sin v = \sin (u+v) - \sin (u-v)\;\;[/imath] we get:
[math]2x_i =[/math][math]\sin \left(a_i-b_1 +b_2-b_3 \right)- \sin \left(a_i-b_1 -b_2+b_3 \right) +[/math][math]\sin \left(a_i- b_2 +b_3-b_1 \right)- \sin \left(a_i-b_2 -b_3+b_1 \right) +[/math][math]\sin \left(a_i-b_3 +b_1-b_2 \right)- \sin \left(a_i-b_3 -b_1 +b_2 \right)[/math][math]=[/math][math]\sin \left(a_i - b_1 + b_2 - b_3 \right) - \sin \left(a_i - b_3 - b_1 + b_2 \right) +[/math][math]\sin \left(a_i - b_2 + b_3 - b_1 \right) - \sin \left(a_i - b_1 - b_2 + b_3 \right) +[/math][math]\sin \left(a_i - b_3 + b_1 - b_2 \right)- \sin \left(a_i - b_2 - b_3 + b_1 \right) =[/math][math]0[/math]