Find the Eigenvalues [[0.65,0.35],[0.45,0.55]]

Problem

[[0.65,0.35],[0.45,0.55]]

Solution

1. Set up the characteristic equation by subtracting λ from the diagonal elements of the matrix A and setting the determinant of A−λ*I to zero.

det(0.65−λ)=0

2. Expand the determinant using the formula for a 2×2 matrix, which is (a−λ)*(d−λ)−b*c

(0.65−λ)*(0.55−λ)−(0.35)*(0.45)=0

3. Multiply the terms to form a quadratic equation in terms of λ

λ2−0.55*λ−0.65*λ+0.3575−0.1575=0

4. Simplify the quadratic equation by combining like terms.

λ2−1.2*λ+0.2=0

5. Solve for λ using the quadratic formula or factoring. Since 1 - 0.2 = 0.8a*n*d \times 0.2 = 0.2d(o)*e*s(n)*o*t*ƒ*i*t,w*e*ƒ*a*c*t*o*r*a*s()\lambda - 1)(\lambda - 0.2) = 0$.

(λ_1)=1

(λ_2)=0.2

Final Answer

λ=1,0.2

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