Question

A mathematical equation for resolution ...

Answer 1

Given the following matrices:

M(δ) = \[\left[\begin{matrix}\exp(kδ/2) & 0 \\ 0 & \exp(-kδ/2)\end{matrix}\right]\] and P(θ) = \[\left[\begin{matrix}\cosθ & sinθ \\ -sinθ & \cosθ\end{matrix}\right]\]

show that

P(-θ)M(δ)P(θ) = \[\left[\begin{matrix}cos(δ/2)+kcos(2θ)sin(δ/2) & ksin(2θ)sin(δ/2) \\ ksin(2θ)sin(δ/2) & cos(δ/2)-kcos(2θ)sin(δ/2)\end{matrix}\right]\]

Answer 2

Well, you know your trig for things like \(\sin\theta\) vs \(\sin-\theta\) and same for cos?

Answer 3

@e.mccormick : Are you referring to sin(-θ) = -sinθ & cos(-θ) = cosθ?

Answer 4

Yes.

So it looks like you need to write out the matrices in order, do some multiplication and simplification.

Answer 5

Yes, I tried to do that, but what stumped me was how it is possible to convert exp(kδ/2) into a trig function (without the exponent), considering that there is no indication of an imaginary power for the exponent (i.e. k≠i) and Euler's formula cannot be applied?

Answer 6

Rotational matrices? Reflection? Hmm... I know some of those use the patterns along the lines of the ones you have.

Answer 7

It's a rotational matrix. But how do you remove the exp & convert it into a trig function in this case?

Answer 8

@mathmale

Answer 9

this is valid for k=i (iota) only imo.

Answer 10

for general k, it won't be possible.

Answer 11

the matrix on the right side has (cosx, sinx) and (-sinx, cosx) as eigenvectors with eigen values (cosd+ksind) and (cosd-ksind), where k=delta/2. These are equal to the given eigenvalues of the original matrix only if k is iota.

Answer 12

@nipunmalhotra93 : OK, but even when treating k=i [i.e. the imaginary constant, √(-1)], I still do not get the solution. For instance, for the second element (i.e. the element in row 1, column 2) of the matrix, I get cos(δ/2)sin2θ [instead of ksin2θsin(δ/2)].

Answer 13

How are you tackling this problem? I mean are you trying to just do the normal multiplication of the 3 matrices on the left to get the matrix on the right? Because that's not what I did.

Answer 14

@nipunmalhotra93 : I find M(δ)P(θ) first, then I map this with P(-θ) to get P(-θ)M(δ)P(θ).

Answer 15

Yeah that's what I meant. Although I haven't tried that, it'd get quite messy imo.

Answer 16

Are you familiar with the concept of eigenvectors?

Answer 17

Yes, I am

Answer 18

you have to show that P(-θ)M(δ)P(θ)=A.
Which is equivalent to showing that M(δ)=P(θ)AP(-θ). Do you agree?

Answer 19

@nipunmalhotra93 : Yes, you're correct

Answer 20

Also, P(-θ) is the rotation matrix with the columns (cosθ,sinθ) and (-sinθ, cosθ). If you prove that these are eigenvectors of A with eigenvalues exp(kδ/2) and exp(-kδ/2) respectively, the above equality'd be proved. (k=i)

Answer 21

^btw I used change of basis here.... from the standard ordered basis {(1,0),(0,1)} to {(cosθ,sinθ),(-sinθ, cosθ)}

Answer 22

@nipunmalhotra93 : Are you saying that for P(θ), the eigenvectors are:

\[\left(\begin{matrix}cosx \\ sinx\end{matrix}\right)\] and \[\left(\begin{matrix}-sinx \\ cosx\end{matrix}\right)\]?

Answer 23

No no... these are the eigenvectors for A.

Answer 24

Actually there is a change of basis here... do you want me to explain that?

Answer 25

@nipunmalhotra93: You can't do a 'change of basis' (to quote your previous post) here, since by doing that [i.e. by converting {(1,0),(0,1)} to {(cosθ,sinθ),(-sinθ, cosθ)} (as what you had stated in your previous post)], you're assuming that θ takes on a set of fixed values, i.e. (more specifically) the GS of θ = 360n° (where n ∈ Z). Also, I don't mean to be rude here, but I don't believe that you know what you're writing (as the evidence from your posts so far seems to portray this). If you're unsure on what you intend to write, please don't include it, as it only creates confusion for others.

Answer 26

wohhhoo... calm down bro XD

Answer 27

and I didn't get your reasoning as to why one can't do a change of basis here.... please explain

Answer 28

@nipunmalhotra93 : You can't do a change of basis since by doing that, you are essentially fixing the values of θ, i.e. you have set cosθ = 1, and sinθ=0, which only occurs when θ = 360n° (where n ∈ Z) [as when you equated the matrix {(1,0),(0,1)} to {(cosθ,sinθ),(-sinθ, cosθ)}].

Answer 29

O_O ..... O_O ummm.... actually θ can be anything. The vectors (cosθ,sinθ),(-sinθ, cosθ) are eigenvectors of the matrix A (the matrix on the rhs). By using these vectors as the basis, A gets converted to the original diagonal matrix (whose non zero entries are the eigenvalues of the given vectors).

Answer 30

@nipunmalhotra93 : I need to log off now - I'll likely return another time (possibly tomorrow).

Answer 31

Solved -

Answer 32

using what?

Answer 33

just simple matrix multiplication & Euler's identity

Answer 34

-_- k