Linear Algebra, 2020a

420 Chapter Five. Similarity
is diagonalizable with λ 1 = 1, λ 2 = 2, and λ 3 = 3 and these are associated
eigenvectors, which make up a basis B.
⎛
⎜
2
⎞ ⎛ ⎞ ⎛
9
⎟ ⃗β 1 = ⎝1⎠ β ⃗ ⎜ ⎟
2 = ⎝4⎠ ⃗ ⎜
2
⎞
⎟
β 3 = ⎝1⎠
0
4
2
The arrow diagram
gives this.
V wrt E3
id
⏐
↓
V wrt B
t
−−−−→
T
t
−−−−→
D
V wrt E3
id
⏐
↓
V wrt B
D = P −1 TP
⎛
⎜
1 0 0
⎞ ⎛
⎟ ⎜
−2 5 −1/2
⎞ ⎛
⎞ ⎛
2 −2 2
⎟ ⎜
⎟ ⎜
2 9 2
⎞
⎟
⎝0 2 0⎠ = ⎝ 1 −2 0 ⎠ ⎝ 0 1 1⎠
⎝1 4 1⎠
0 0 3 −2 4 1/2 −4 8 3 0 4 2
The bottom line of the diagram has t(⃗β 1 )=1 · ⃗β 1 , etc. That is, the action on
the basis is this.
t−1
⃗β 1 ↦−→ ⃗0
t−2
⃗β 2 ↦−→ ⃗0
t−3
⃗β 3 ↦−→ ⃗0
Here is how the top line of the arrow diagram represents the first of those three
actions
(T − 1 · I)⃗β 1 = ⃗0
⎛
⎞ ⎛
1 −2 2
⎜
⎟ ⎜
2
⎞ ⎛
⎟ ⎜
0
⎞
⎟
⎝ 0 0 1⎠
⎝1⎠ = ⎝0⎠
−4 8 2 0 0
(of course, the representation of ⃗β 1 with respect to the standard basis is itself).
This section observes that some matrices are similar to a diagonal matrix.
The idea of eigenvalues arose as the entries of that diagonal matrix, although
the definition applies more broadly than just to diagonalizable matrices. To find
eigenvalues we defined the characteristic equation and that led to the final result,