Eigenvectors problem - almost solved, need a little guidance

[abayaz]

Hi guys,

I'm trying to solve the attached problem.

I work out the first few powers of A by hand

Then I take the eigenvectors provided and put them in a new matrix E, as two columns.

Then I take the inverse of that matrix E (using an online inverse matrix calculator)

Then I calculate E^-1 A E , which is a diagonal matrix - as expected. I then calculate the square of this.... and now I'm kind of lost as to how to use this to get A²

 

[ksdhart2]

You're nearly there. You just need to take the very last step. You've constructed a diagonal matrix based on the eigenvalues. Unfortunately, you didn't share with us what this matrix is, so I'll just have to assume for now that you've done this part correctly. In any case, we need to be able to work with this diagonal matrix, so let's give it a name. Let \(D = E^{-1} \: A \: E\).

You say you calculated \(D^2\) to help find \(A^2\). That's a great idea - start simple and build up from there. Let's look symbolically at what that step gave us:

\(D^2 = (E^{-1} \: A \: E)^2 = (E^{-1} \: A \: E) \times (E^{-1} \: A \: E)\)

Hmm... that looks a bit ugly. Can we simplify that any? Note that \(E \: E^{-1} = I\) and:

\( D^2 = E^{-1} \: A \: E \: E^{-1} \: A \: E = E^{-1} \: A^2 \: E \)

Oh, now it looks like we're going somewhere! We want to rearrange this formula so that \(A^2\) is the subject. How might you do that? Once you have that, don't forget to check to make sure the answer you got makes sense. If it does, try following the same process for \(A^3\). Does that result also make sense? What does that suggest about a formula for \(A^n\)?

 

[HallsofIvy]

In fact, for any positive integer, n, \(\displaystyle A^n= (E^{-1}DE)^n(E^{-1}DE)(E^{-1}DE)\cdot\cdot\cdot (E^{-1}DE)(E^{-1}DE)= E^{-1}D(EE^{-1})D(E^{-1}E)D(E^{-1}E)\cdot\cdot\cdot D(E^{-1}E)DE= E^{-1}D^nE\)