sh3lsh
  • sh3lsh
Linear Algebra fun times. Described with Latex below.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sh3lsh
  • sh3lsh
If 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
  • sh3lsh
Also, 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
  • Empty
I think to be diagonalizable you just have to have a nonzero determinant. I could be wrong though.

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Empty
  • Empty
I 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
  • sh3lsh
Eh? the way you calculate eigenvectors is to do to \[\det (A- \lambda I_{n})\]
Empty
  • Empty
I 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://s-mat-pcs.oulu.fi/~mpa/matreng/eem4_2-3.htm Ok, it's a start
sh3lsh
  • sh3lsh
(add an = 0 to that)
sh3lsh
  • sh3lsh
Nonetheless was my way of calculating A^5 correct?
Empty
  • Empty
For 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
  • sh3lsh
Hmm. That person left and deleted their response! But thanks @Empty! I was just checking. You're awesome.
Empty
  • Empty
Yeah weird I don't know why #_# I thought they had some good responses
Empty
  • Empty
Diagonalization 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
  • sh3lsh
Interesting. My professor doesn't tell us things like this! I appreciate this!
Empty
  • Empty
Yeah if you wanna learn anything about linear algebra just ask I'd love to share and introduce some fun stuff

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