1. Feb 12, 2012

### sidinsky

M = (a c)
(c b)

Sorry for the double sets of brackets, its all in one. I'll also show as far as i got below:

[a-λ c] => (a-λ)(b-λ) - c^2 = λ^2 + (-a-b)λ + (ab-c^2) =0
[c b-λ] =>

then using the quadratic formula: λ = [-(-a-b) +/- Sqrt{(-a-b)^2 - 4(1)(ab-c^2)}]/ 2

then after some algebra I got stuck: λ = [(a+b) +/- Sqrt{a^2 + 2ab + b^2 -4ac^2}]/2

λ = [(a+b) +/- Sqrt{a^2 - 2ab + b^2 + c^2}]/2 ==>> THIS IS WHERE I GOT STUCK. PLEASE HELP

2. Feb 12, 2012

### tiny-tim

welcome to pf!

hi sidinsky! welcome to pf!

(have a square-root: √ and a ± and try using the X2 button just above the Reply box )
looks ok

that gives you the two eigenvalues,

so now use the standard techniques to find an eigenvector for each

3. Feb 13, 2012

### sidinsky

is there any way this expression can be simplified further? I am sorta having trouble with the algebra :S

4. Feb 14, 2012

### tiny-tim

i don't think so

how far have you got?​

5. Feb 14, 2012

### sidinsky

well I managed to clean up the expression inside the sqrt a bit to: [(a-b)2 + c 2 ]/2

a2 -2ab + b2 + c2 = (a-b)2 + c2

λ1 = (a+b) + sqrt{(a-b)2 +c 2}$/2$

and λ2 = (a+b) - sqrt{(a-b)2 + c2}$/2$

now these expressions for Lambda are too difficult for me to use to solve for eigenvectors