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Mathematics 8 Online
OpenStudy (anonymous):

Let v be an eigenvector of the matrix M, with corresponding eigenvalue λ. Show that v is also an eigenvector of the matrix M^2 with corresponding eigenvalue λ^2. Your kind assistance is always appreciated. Many Thanks...

OpenStudy (anonymous):

I mean M

OpenStudy (anonymous):

so what is M^2

OpenStudy (anonymous):

Something happened to my equation

OpenStudy (anonymous):

Let L = the diagonal matrix with the eigenvalues

OpenStudy (anonymous):

I have no idea ;( i saw it briefly then it disappeared. :-)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

v L v^-1 = M v L v^-1 v L v^-1 = M M

OpenStudy (anonymous):

Whaddya got?

OpenStudy (anonymous):

is v the eigenvector and are you multiplying it with - landa v_1 ?

OpenStudy (anonymous):

Yes By definition v lambda v^-1 = M

OpenStudy (anonymous):

okay, and M is a normal matrix correct?

OpenStudy (anonymous):

normal? I think it has to be square.

OpenStudy (anonymous):

sorry but what is v^-1 ?

OpenStudy (anonymous):

yes i mean square 2x2 , 3x3 nxn

OpenStudy (anonymous):

You get your eigenvectors and put em in a matrix and then invert it.

OpenStudy (anonymous):

right?

OpenStudy (anonymous):

Sure

OpenStudy (anonymous):

That's the whole idea v L v^-1 = M

OpenStudy (anonymous):

and is that all two or all three at a time?

OpenStudy (anonymous):

So you can chain them , that's what's great.

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

The v v^-1 all cancel.

OpenStudy (anonymous):

right

OpenStudy (anonymous):

So then we have M^2 = v L^2 v^-1

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

eigenvectors right there

OpenStudy (anonymous):

squared eigenvalues in a diagonal matrix

OpenStudy (anonymous):

I might have to read up on this a bit more...under what specific topic can I find this ? Is it under properties of eigenvectors/eigenvalues?

OpenStudy (anonymous):

Sure. I have trouble remembering which lecture, but I learned this from Gil Strang's MIT ocw course in linear algebra. He's the best teacher I have ever seen.

OpenStudy (anonymous):

was that the online course?

OpenStudy (anonymous):

Yep

OpenStudy (anonymous):

i will have a look, do u think that khanacademy covers this?

OpenStudy (anonymous):

I'm pretty sure he does. Advantage of Khan is short and sweet. Disadvantage is he is fragmented and sometimes makes mistakes. Your pref.

OpenStudy (anonymous):

I think it might fall under the definition - similar matrices, I must have a look... I love MIT though, and its great when you have loads of time...

OpenStudy (anonymous):

I will get back to you tomorrow with chapter. Are you good with basic facts about Linear Algebra?

OpenStudy (anonymous):

I'm getting better, but need to revise here and there....I'm finishing 1st year engineering so I will have to be good with basic facts ...hehehe

OpenStudy (anonymous):

Yes. good luck.

OpenStudy (anonymous):

I will watch out for you tomorrow. Will you msh me the chapter please. Much Appreciated...thank you . for your kind help

OpenStudy (anonymous):

Tell you what, I will explain Linear Algebra You can explain Fourier series and Laplace Transform. !!!

OpenStudy (anonymous):

hehehe...At the end of year 2!!! Thats 12 months from now!

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