Question

how to find a power of a matrix that is not diagonizable?? for example the following matrix 3x3 matrix A has only 2 eigenvalue and 2 eigenvectors how can i find power of this matrix....
A= (0 1 0; 0 0 1; 4 -8 5)
here colon indicates end of row

Answer 1

i was trying to find its diagonal matrix

Answer 2

then i found it does not have n independent eigen vectors

Answer 3

so i don know other way to find its nth power

Answer 4

I'm not exactly sure what you're trying to do... just raise the matrix to an arbitrary power, or what?

Answer 5

i need to calculate the nth power of the following matrix....

Answer 6

http://www.maths.lse.ac.uk/Personal/martin/fme4a.pdf
Maybe this will help

Answer 7

Neuro, is this a theoretical question or one where you just need to actually calculate the nth power and have a result?

Answer 8

@swissgirl sorry, but i already said this matrix is not diagonizable....so that does not help...

Answer 9

@mathteacher1729 .... ya....i need a result but that can be in the form of variable n.....if not a good explanantion would be ok as well...

Answer 10

it is part of a big question so i don want to put everything here....but all i need is to find A power n and A is exactly as i put into example here

Answer 11

You can try and do a diagonalization and then start taking successive powers of the (almost diagonal) "D" matrix. You'll spot a pattern, but it's not pretty.

To see the full result , go to wolframalpha and type

matrix power[{{0,1, 0},{ 0, 0, 1},{ 4, -8, 5}},n]

Answer 12

i know abt diagonalization but here the problem is, A has only 2 eigenvalue and 2 eigen vector so i can not diagonalize it

Answer 13

@mathteacher1729 does that have to do with "similar" matrixes?

Answer 14

this matrix has 3 distinct eugene vectors

Answer 15

what makes you say it doesnt have n distinct eugenes?

Answer 16

i can find 1 1 1 and 1 2 4

Answer 17

@neuro11 have you learned about jordan factorizations yet? My next thought would be to find the Jordan Normal Form: http://www.ms.uky.edu/~lee/amspekulin/jordan_canonical_form.pdf

Answer 18

it has 3 repeated values 1,1,1 correct?

Answer 19

@amistre64 yes, this has to do with similarity. :)
@neuro11 you can check your work by using wolfram alpha and typing

{{0,1, 0},{ 0, 0, 1},{ 4, -8, 5}}

Answer 20

linear class stopped right before the orthonormal stuff .... never did get to similars :/

Answer 21

@mathteacher1729 i will check in wolfram

Answer 22

@amistre64 i have eigenes 1 and 2 and vector (1 1 1) and (1 2 4).... i will chk in wolfram

Answer 23

i inputed the wrong data into the wolf to begin with,
http://www.wolframalpha.com/input/?i=%7B+%7B0%2C+1%2C+0%7D%2C%7B0%2C+0%2C+1%7D%2C%7B+4%2C+-8%2C+5%7D%7D

should help

Answer 24

i did not know that for a real double root (like 2 in this case) if no vector is found it can be written as (0,0,0)