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windsylph
: Linear Algebra: (Diagonalization) Let A = {{1,1},{0,1}}. Use the definition of the matrix exponential to compute e^A.
A = \[\left[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right]\]
Write A as S + N, where S is the identity, and N is the nilpotent matrix with only a 1 in the 1,2 position.
Okay, here's what I got before I posted this question: {{e, 1+ 2/2! + 3/3! + ... + n/n!},{0,e}}. The book says {{e,e},{0,e}}, but I don't get how 1+ 2/2! + 3/3! + ... + n/n! = e, lest I'm wrong on the row 1 col 2 entry.
more scanning, one sec.
the idea is if you matrix isnt diagonalizable, you want to write it as the sum of a matrix that is, and a matrix that is nilpotent, which means that after some finite power, the matrix will be 0. Also, these matrices must commute.
hm, I really can't use the fact that A = S+N since our haven't gone over nilpotent matrices yet (or we won't at all), so here's what I did, haha I don't know how to go from here to {{e,e},{0,e}}.
oh cool, so you came up with an idea of what A^n might be by observation, and used that, very clever :) To make that method a little more solid, prove your guess for A^n by Induction. Then nobody will have any problem with that solution :)
Something like this, its short, but it gives your argument more foundation :) very nice observation.
oops forgot some 0's in there <.<
i hope they teach you about nilpotent matrices =/ thats really important when calculating matrix exponentials. In general, you might not be able to guess a formula for A^n if it isnt diagonalizable. thats why you need to be able to write it as S + N, where S is diagonalizable, and N is nilpotent, and S and N commute. If you are interested in that sorta stuff, look up the Jordan Canonical Form.