Practice Final Question #1 Explanation
Formulate the matrix A- λI to solve for the eigenvalues. The eigenvalues are λ=2, and λ=-1 with an algebraic multiplicity of 2 since you find that the det(A- λI)=( λ-2)( λ+1)^{2} . These eigenvalues placed along the diagonal is the matrix D. Find eigenvectors for each corresponding eigenvalue by plugging in each corresponding eigenvalue into A- λI and row reducing that matrix in order to find the eigenvectors. For λ=2, the eigenvector is (1 1 1 ) (In a column). For λ=-1, you get the eigenvectors (-1 1 0), (-1 0 1).
brookewenig: May 29, 2015, 8:58 p.m.
I got the same values for the eigenvalues as you, but I did not reach the same factorization as you. \\ How did you get to det(A- λI)=( λ-2)( λ+1)^{2}? I was only able to get to det(A- λI)= (λ)(λ^2 - 3) = 2, which after plugging in numbers yields λ = 2, -1. If I had factored the same way you did, I would not have needed to guess values for λ. Any help is appreciated!
tedallen: May 30, 2015, 12:59 p.m.
If you assume that one of the roots is repeated, you can represent the equation as (λ^2+Aλ+B)(λ+C)=0. Since you know the coefficients for λ^3, λ^2, λ, and 1, you get the equations BC = 2; C + A = 0; and (A^2)/4 = B. Plug in the second two into the first, and you'll get values for A,B,&C.
alexwang0518: June 1, 2015, 7:58 p.m.
@Brookewenig, Factoring a cubic can be hard. You can use factors of the constant to guess the eignvalues. For instance you should get the determinant function to be: - λ^3 + 3 λ + 2. 2 has factors: 1, -1, 2, -2. Now, plug in each factor into the determinant function to see which ones will equal zero. Turns out 2 and -1 make it zero so then you can figure out how to factor the cubic. Hopefully, the test will just have an easier cubic / no cubic at all.
brookewenig: June 2, 2015, 7:52 a.m.