Finding Eigenvalues for Tridiagonal Matrix - Showing Orthogonality

[eckiller]

I have the tridiagonal matrix (which comes from the backward Euler scheme)

A =
[ 1+2M - M 0 ... ]
[ -M 1+2M 0 ... ]
[ ... ]
[ -M 1+2M ]


I am given that the eigenvectors are v_k = [ sin(pi*k*x_j ] for j = 1:J-1 and k = 1:J-1

I am to find the eigenvalues L_k.

I tried just writing A * v_k = L_k * v_k

Then doing the multiplication on the left hand side I can get a relationship from one equation:

-M*sin(sin(pi*k*x_j-1) + (1+2M)sin(pi*k*x_j) -M*sin(pi*k*x_j+1) = L_k * sin(pi*k*x_j)

I get stuck here. Is this the right approach? Is there someway to solve for L_k?


Also: I am to show the v_k are orthogonal. Would I use dot product or the integral inner product to show this?

Thanks for any help.

 

[CarlB]

I get stuck here. Is this the right approach? Is there someway to solve for L_k?

You've got the right answer. If you think the instructor wants you to reduce it to simplest form, then look up the trig identity for sin(A+B) to bring those nasty sines into more equal terms.

Also: I am to show the v_k are orthogonal. Would I use dot product or the integral inner product to show this?
Thanks for any help.

I would guess that they're wanting you to use the integral inner product. That is, they want you to integrate over x the dot product. But the nature of dot produts should give you that the dot product itself is zero for all x so your integral will be trivial.

Good luck. And if you want your questions answered faster, try learning LaTex and making your questions pretty.

Carl