Question

Elementary linear algebra question:

If \( A = \begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix} \) then, find \( I+A+A^2+ \cdots \infty \)

Answer 1

haha how wld u solve this?

Answer 2

Through magic lol, lets bring Turing here.

Answer 3

omg tghis cat is getting menervous. I hate its eyes

Answer 4

Your definition of elementary is pretty bold fool

Answer 5

foool is bold

Answer 6

can anybody help me on some problems nobody is answering them... plz :))

Answer 7

lol, taken from a high school level book that's elementary, isn't ? :)

Answer 8

Useful fact
\[ \frac{1}{1-\lambda} = 1 + \lambda + \lambda^2 + .... \] provided \( |lambda| < 1 \)

Answer 9

eigenvalues?

Answer 10

or you just mean the sum of the geometric series?

Answer 11

Yeah, why else James use \( \lambda \)? He could have used \(x\)

Answer 12

yes, if you can find a diagonal matrix D such that
\[ A = P^{-1}DP \]
then this sum, if it exists is
\[ P^{-1}(I + D + D^2 + ....)P \]
and those terms are easy to calculate.

Answer 13

why is this in a hs textbook??????

Answer 14

I'm really not that good at linear algebra, I don't know that theorem james put up, and I'm trying to put the pieces together

Answer 15

Turing : http://en.wikipedia.org/wiki/Diagonalizable_matrix

Answer 16

Yeah this rings a bell
gotta totally relearn in though

Answer 17

Sweet :)

Answer 18

sweet????? that didnt fit in lol

Answer 19

lol, sour then :)

Answer 20

\[\frac{1}{6}\left[\begin{matrix}3 & -2 \\ -3 & 0\end{matrix}\right]\]

Answer 21

is that the answer?

Answer 22

I keep getting an ugly eigenvalue

Answer 23

*valueS

Answer 24

Somewhere in Lecture 23 of Strang's LA he said
(I-At)^(-1)= I + At + (At)^2 +...

Answer 25

phi is right!

Answer 26

At ?

Answer 27

transpose of A?

Answer 28

t is a variable
He started by defining e^(At) where A is a matrix
then did a btw

Answer 29

hmm, I'll have to do OCW linear algebra after multivariable

Answer 30

He's very good.

Answer 31

Turing never gets enough of math lol

Answer 32

So here you are doing \( (I-A)^ {-1} \) ?

Answer 33

Yes, phi's approach is shorter and more elementary. In short, he wins the prize for this question.

Answer 34

@JamesJ: Clearly it is, here is however a elementary proof that I came up with,

\[ I+A+A^2+ \cdots \infty = X (say) \]
\[ \implies A+A^2+A^3 \cdots \infty = AX \]

Substracting,

\[ X(I-A) = I \implies X = (I-A)^{-1} \space \space (QED) \]