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

[Violette]

prove this Det is equal to 0 ? I have thought of using cos(a-b)=cosacosb+sinasinb but ended up having no ideas how to sovle this problem.

 

[khansaheb]

View attachment 36612
prove this Det is equal to 0 ? I have thought of using cos(a-b)=cosa*cosb + sina*sinb but ended up having no ideas how to solvl this problem.

Were you given any other relationship between 'a' & 'b' & 'n' or restriction on their ranges? Can n=1?

Have you given us the whole problem as it was presented to you to you?

 

[topsquark]

View attachment 36612
prove this Det is equal to 0 ? I have thought of using cos(a-b)=cosacosb+sinasinb but ended up having no ideas how to sovle this problem.

D is not 0, even for a 2 x 2 determinant. Why do you think it should be 0?

-Dan

 

[khansaheb]

D is not 0, even for a 2 x 2 determinant. Why do you think it should be 0?

-Dan

That became clear to me - even for 1x1 matrix

 

[Violette]

Were you given any other relationship between 'a' & 'b' & 'n' or restriction on their ranges? Can n=1?

Have you given us the whole problem as it was presented to you to you?

Were you given any other relationship between 'a' & 'b' & 'n' or restriction on their ranges? Can n=1?

Have you given us the whole problem as it was presented to you to you?

The problem didn't give any relationship between 'a' & 'b' & 'n' or restriction on their ranges. We have n in there like n can be infinity. Yes I have given th whole problem exactly like it was presented to me it said " Chứng minh định thức D = 0" translate to English means " Prove the determinant equals to 0"

 

[Violette]

D is not 0, even for a 2 x 2 determinant. Why do you think it should be 0?

-Dan

That's why I am stuck here because I have no idea why it should be 0. Maybe somehow when we factor out all it will lost all?
(using row operations)

 

[topsquark]

That's why I am stuck here because I have no idea why it should be 0. Maybe somehow when we factor out all it will lost all?
(using row operations)

Nope. For the 2 x 2 it gives
[imath]D = sin(a_1-b_1) \, sin(a_2-b_2)[/imath]

if I remember correctly.

-Dan

 

[Violette]

Nope. For the 2 x 2 it gives
[imath]D = sin(a_1-b_1) \, sin(a_2-b_2)[/imath]

if I remember correctly.

-Dan

hmn so the question is unsolvable. How about finding (a,b) that makes D equals 0 ? ^^

 

[khansaheb]

How about finding (a,b) that makes D equals 0 ?

How about it?

Start working and show us your work.

 

[topsquark]

Nope. For the 2 x 2 it gives
[imath]D = sin(a_1-b_1) \, sin(a_2-b_2)[/imath]

if I remember correctly.

-Dan

Correction: For the 2 x 2
[imath]D = sin(a_1-a_2) \, sin(b_1-b_2)[/imath]

I didn't prove it, but for 3 x 3 and 4 x 4, D does equal 0, so the problem should have stated that n > 2.

-Dan

 

[blamocur]

The problem didn't give any relationship between 'a' & 'b' & 'n' or restriction on their ranges.

Are you sure it did not specify [imath]n \geq 3[/imath] ?

 

[blamocur]

Correction: For the 2 x 2
[imath]D = sin(a_1-a_2) \, sin(b_1-b_2)[/imath]

I didn't prove it, but for 3 x 3 and 4 x 4, D does equal 0, so the problem should have stated that n > 2.

-Dan

I did not notice your post until posting my reply to the OP. I cannot prove it either, but my symbolic script shows 0 for [imath]3\leq n \leq 5[/imath], and my numerical script shows zeros for [imath]3 \leq n < 100[/imath].

 

[blamocur]

Unless I am missing something, it should be enough to prove that the determinant is 0 for [imath]n=3[/imath], after which one can apply Laplace expansion.

 

[khansaheb]

Det is equal to 0

This proposition is not true - as presented.

 

[blamocur]

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].

 

[Steven G]

The problem didn't give any relationship between 'a' & 'b' & 'n' or restriction on their ranges. We have n in there like n can be infinity. Yes I have given th whole problem exactly like it was presented to me it said " Chứng minh định thức D = 0" translate to English means " Prove the determinant equals to 0"

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

 

[Violette]

Correction: For the 2 x 2
[imath]D = sin(a_1-a_2) \, sin(b_1-b_2)[/imath]

I didn't prove it, but for 3 x 3 and 4 x 4, D does equal 0, so the problem should have stated that n > 2.

-Dan

Can you give me a brief instruction on how did you analyse D = 0 for 3x3 and 4x4 hmn n > 2

 

[Violette]

yes maybe the book want us to discover it ourselve

Are you sure it did not specify [imath]n \geq 3[/imath] ?

 

[Violette]

what you meant by saying numerical script and symbolic script?

I did not notice your post until posting my reply to the OP. I cannot prove it either, but my symbolic script shows 0 for [imath]3\leq n \leq 5[/imath], and my numerical script shows zeros for [imath]3 \leq n < 100[/imath].

 

[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].

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