x = [ x1 , x2 ] be a unit eigen vector when
√(x1^{2} + x2^{2} ) = 1
So ....(x1^{2} + x2^{2} = 1)
Ax = ∆x , where ∆ = eigen value
= x'Ax = x'∆x (here ' represent transpose)
= ∆x'x ( since ∆ can be think as scalar value)
= ∆ [x1,x2]' [x1,x2]
= ∆ [ x1^{2} + x2^{2} ]
= ∆(1) (since unit eigen vector)
= ∆
From this derivation we can say that max value of x'Ax is the maximum eigen value..
Just find the eigen values which one is max is the ans..
So Max . vakue of x'Ax= ∆ =( 5 + √5) /2.
http://math.stackexchange.com/questions/1553046/what-is-the-maximum-value-of-xt-ax