A community for students.
Here's the question you clicked on:
 0 viewing
sh3lsh
 one year ago
Linear Algebra fun times.
Described with Latex below.
sh3lsh
 one year ago
Linear Algebra fun times. Described with Latex below.

This Question is Closed

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1If one has a diagonalizable matrix A, one can reduce into \[S^{1}AS=D\] So, if we wanted to get A^5, would we just do this: \[S D^{5} S^{1}\]

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Also, if we have a matrix A, when can we tell if it is diagonalizable or not? (just point me to a link if adequate)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I think to be diagonalizable you just have to have a nonzero determinant. I could be wrong though.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I don't think there's a difference between having a complete set of linearly independent eigenvectors and having a nonzero determinant, they're essentially the same condition I believe.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Eh? the way you calculate eigenvectors is to do to \[\det (A \lambda I_{n})\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I think it must not only have a nonzero determinant but also have distinct eigenvalues in order to be diagonalizable. I'm sorta shaky on that, but I checked and found this: http://smatpcs.oulu.fi/~mpa/matreng/eem4_23.htm Ok, it's a start

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Nonetheless was my way of calculating A^5 correct?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2For 2x2 matrices, the characteristic equation will be: \[0=\lambda^2  tr(A) \lambda + det(A)\] So yes you definitely will use the determinant to calculate eigenvalues, although that's not really what I was saying to begin with anyways, I was just using it as a test to see if it had n eigenvalues. @sh3lsh Yes that way is correct! This is how we can even calculate things like noninteger powers of matrices or weird things like \(e^A\) with the Taylor series representation.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Hmm. That person left and deleted their response! But thanks @Empty! I was just checking. You're awesome.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah weird I don't know why #_# I thought they had some good responses

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Diagonalization is also useful for transforming things like ellipses into circles, because these \(S\) matrices will transform the vectors multiplying the quadratic form so that you don't have any cross terms, only the diagonal squared terms. Diagonalization is just an awesome thing to have really.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.1Interesting. My professor doesn't tell us things like this! I appreciate this!

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah if you wanna learn anything about linear algebra just ask I'd love to share and introduce some fun stuff
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.