- #1
sponsoredwalk
- 533
- 5
I just can't figure out how you arrive at having a diagonal matrix consisting of Jordan blocks.
Going by Lang, a vector is (A - λI)-cyclic with period n if (A - λI)ⁿv = 0, for some n ∈ℕ.
It can be proven that v, (A - λI)v, ..., (A - λI)ⁿ⁻¹ are linearly independent, & so
{v, (A - λI)v, ..., (A - λI)ⁿ⁻¹} forms a basis, called the Jordan basis, for what is now known
as a cyclic vector space.
Furthermore, for each (A - λI)ⁿv we have that (A - λI)ⁿv = (A - λI)ⁿ⁺¹v + λ(A - λI)ⁿv.
Now for the life of me I just don't see how the matrix associated to this basis is a matrix
consisting of λ on the diagonal & 1's on the superdiagonal.
But assuming that works, I don't see how taking the direct sum of cyclic subspaces can
be represented as a matrix consisting of matrices on the diagonal.
Basically I'm just asking to see explicitly how you form the matrix w.r.t. the Jordan basis
& to see how you form a matrix representation of a direct sum of subspaces, appreciate
any & all help.
Going by Lang, a vector is (A - λI)-cyclic with period n if (A - λI)ⁿv = 0, for some n ∈ℕ.
It can be proven that v, (A - λI)v, ..., (A - λI)ⁿ⁻¹ are linearly independent, & so
{v, (A - λI)v, ..., (A - λI)ⁿ⁻¹} forms a basis, called the Jordan basis, for what is now known
as a cyclic vector space.
Furthermore, for each (A - λI)ⁿv we have that (A - λI)ⁿv = (A - λI)ⁿ⁺¹v + λ(A - λI)ⁿv.
Now for the life of me I just don't see how the matrix associated to this basis is a matrix
consisting of λ on the diagonal & 1's on the superdiagonal.
But assuming that works, I don't see how taking the direct sum of cyclic subspaces can
be represented as a matrix consisting of matrices on the diagonal.
Basically I'm just asking to see explicitly how you form the matrix w.r.t. the Jordan basis
& to see how you form a matrix representation of a direct sum of subspaces, appreciate
any & all help.