Linear Algebra, 2020a

Section IV. Jordan Form 453
The diagonalizable case is where t has n distinct eigenvalues λ 1 , ..., λ n ,
that is, where the number of eigenvalues equals the dimension of the space. In
this case there is a basis 〈⃗β 1 ,...,⃗β n 〉 where ⃗β i is an eigenvector associated with
the eigenvalue λ i . The example below has n = 3 and the three eigenvalues are
λ 1 = 1, λ 2 = 3, and λ 3 =−1. It shows the canonical representative matrix and
the associated basis.
⎛
⎜
1 0 0
⎞
⎟
T 1 = ⎝0 3 0 ⎠
0 0 −1
⃗β 1
t−1
↦−→ ⃗0
⃗β 2
t−3
↦−→ ⃗0
⃗β 3
t+1
↦−→ ⃗0
One diagonalization example is Five.II.3.21.
The case where t has a single eigenvalue leverages the results on nilpotency
to get a basis of associated eigenvectors that form disjoint strings. This example
has n = 10, a single eigenvalue λ = 2, and five Jordan blocks.
⎛
⎞
2 0 0 0 0 0 0 0 0 0
1 2 0 0 0 0 0 0 0 0
0 1 2 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0
T 2 =
0 0 0 1 2 0 0 0 0 0
0 0 0 0 1 2 0 0 0 0
0 0 0 0 0 0 2 0 0 0
0 0 0 0 0 0 1 2 0 0
⎜
⎟
⎝0 0 0 0 0 0 0 0 2 0⎠
0 0 0 0 0 0 0 0 0 2
⃗β 1
t−2
↦−→ ⃗β 2
t−2
↦−→ ⃗β 3
t−2
↦−→ ⃗0
⃗β 4
t−2
↦−→ ⃗β 5
t−2
↦−→ ⃗β 6
t−2
↦−→ ⃗0
⃗β 7
t−2
↦−→ ⃗β 8
t−2
↦−→ ⃗0
⃗β 9
t−2
↦−→ ⃗0
⃗β 10
t−2
↦−→ ⃗0
We saw a full example as Example 2.5.
Theorem 2.8 below extends these two to any transformation t: C n → C n .
The canonical form consists of Jordan blocks containing the eigenvalues. This
illustrates such a matrix for n = 6 with eigenvalues 3, 2, and −1 (examples with
full compuations are after the theorem).
⎛
⎞
3 0 0 0 0 0
1 3 0 0 0 0
0 0 3 0 0 0
T 3 =
0 0 0 2 0 0
⎜
⎟
⎝0 0 0 1 2 0 ⎠
0 0 0 0 0 −1
⃗β 1
t−3
↦−→ ⃗β 2
t−3
↦−→ ⃗0
⃗β 3
t−3
↦−→ ⃗0
⃗β 4
t−2
↦−→ ⃗β 5
t−2
↦−→ ⃗0
⃗β 6
t+1
↦−→ ⃗0
It has four blocks, two associated with the eigenvalue 3, and one each with 2
and −1.