Problem with showing properties of linear transformation Let F:K3→K3 be a linear transformation, such that 1. (F−λI)(F−μI)≠0, 2. (F−μI)2≠0, 3. (F−λI)(F−μI)2=0. I want to show that 1. λ,μ are eigenvalues of F, 2. ker(F−μI)⊆Im(F−μI), 3. F has in a certain basis matrix (λ000μ100μ)

Question

Problem with showing properties of linear transformation  Let F:K3→K3 be a linear transformation, such that  1. (F−λI)(F−μI)≠0,  2. (F−μI)2≠0,  3. (F−λI)(F−μI)2=0.  I want to show that  1. λ,μ are eigenvalues of F,  2. ker(F−μI)⊆Im(F−μI),  3. F has in a certain basis matrix  (λ000μ100μ)

Justice Jacobson · Accepted Answer

I will assume λ≠μ.  Note that the factors F−kI commute for all k.  1. Let v be a vector v∈K3 such that  (F−λI)(F−μI)v=u≠0. Then:  (F−μI)u=(F−λI)(F−μI)2v=0  so u is an eigenvector for μ.   Analogously, let w be a vector w∈K3 such that (F−μI)2w=z≠0.  Then:  (F−λI)z=(F−λI)(F−μI)2w=0  so z is an eigenvector for λ.  2. In part 3. we can see that the dimension of ker(F−μI) is 1, otherwise F would be diagonalizable. So u∈ker(F−μI). But u=(F−μI)vtextwherev=(F−λI)v thus u∈Im(F−μI).  3. The polynomial (F−λI)(F−μI)2 from statement 3 is the characteristic polynomial, because it has degree three, and it does not vanish when removing any of its factors. Thus μ has multiplicity 2 and the diagonal of the Jordan form of F is (λ,μ,μ)T. But if F were diagonalizable, then (F−λI)(F−μI)=0, which is in contradiction with statement 1.

Problem with showing properties of linear transformation Let F:K^{3} \rightarrow K^{3} be

Answered question

Answer & Explanation

New Questions in Integral Calculus