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The Eigenvalues and eigenvectors of a 2x2 matrix

by LydiaSylar
Tags: eigenvalues, eigenvectors, linear algebra
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LydiaSylar
#1
Nov15-09, 11:25 AM
P: 1
1. The problem statement, all variables and given/known data

Let B = (1 1 / -1 1)
That is a 2x2 matrix with (1 1) on the first row and (-1 1) on the second.

2. Relevant equations



3. The attempt at a solution

A)

(1 1 / -1 1)(x / y) = L(x / y)

L(x / y) - (1 1 / -1 1) (x / y) = (0 / 0)

({L - 1} -1 / 1 {L-1}) (x / y) = (0 / 0)

Det (LI - B) = ({L - 1} -1 / 1 {L-1}) = 0

({L - 1} {L-1}) - (1)(-1)

L^2 -2L +2 = 0

L= 1 - i
= 1+i

So when L = 1-i

({1 -i - 1} -1 / 1 {1 -i -1})

(-i -1 / 1 -i)

-ix - y = 0
x - iy = 0

let x = t

t - iy = 0
y = t/i

Im not sure if that even makes sense. Or how I would continue.

B) Write the eigenvalues L of B in the form w = re^i(theta)

If someone could just give me a little nudge in the right direction for this one because I dont even know where to start.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Dick
#2
Nov15-09, 12:27 PM
Sci Advisor
HW Helper
Thanks
P: 25,251
You are actually doing pretty well. You have the two eigenvalues right, and you've shown that an eigenvector of 1-i is given by x=t, y=t/i=(-it) for any nonzero value of t. That makes it t*(1 / -i). Now just do the same thing for 1+i. For the second part e^(i*theta)=cos(theta)+i*sin(theta). For a complex number L, the 'r' will be |L|. So L/|L|=cos(theta)+i*sin(theta). Just match up the real and imaginary parts and find theta.


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