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 $$3≤ n < 100$$ 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 $a$ and $b$, and they all yielded determinant of zero for $n > 2$.

 

[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 $v$ we have $Mv=0$ or $vM=0$, and $v \neq 0$ then $\det(M) = 0$ ?

 

[Steven G]

What's the difference? why n cannot be infinity ( according to blamocur n value is bounded between $$3≤ n < 100$$ 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 $n=3$ it is not too difficult to construct a vector $v_0$ such that $v_0 M = 0$.

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
$\begin{pmatrix} cos(a_1 - b_1) & cos(a_2 - b_1) & cos(a_3 - b_1) \\ cos(a_1 - b_2) & cos(a_2 - b_2) & cos(a_3 - b_2) \\ cos(a_1 - b_3) & cos(a_2 - b_3) & cos(a_3 - b_3) \end{pmatrix}$

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
$\begin{pmatrix} cos(a_1 - b_1) & cos(a_2 - b_1) & cos(a_3 - b_1) \\ cos(a_1 - b_2) & cos(a_2 - b_2) & cos(a_3 - b_2) \\ cos(a_1 - b_3) & cos(a_2 - b_3) & cos(a_3 - b_3) \end{pmatrix}$

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

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