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Linear Algebra, 2020a

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Section III. Nilpotence 431<br />

2.9 Example The differentiation map d/dx: P 2 → P 2 is nilpotent of index three<br />

since the third derivative of any quadratic polynomial is zero. This map’s action<br />

is described by the string x 2 ↦→ 2x ↦→ 2 ↦→ 0 and taking the basis B = 〈x 2 ,2x,2〉<br />

gives this representation.<br />

⎛<br />

⎜<br />

0 0 0<br />

⎞<br />

⎟<br />

Rep B,B (d/dx) = ⎝1 0 0⎠<br />

0 1 0<br />

Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones.<br />

2.10 Example With the matrix ˆN from Example 2.5, and this four-vector basis<br />

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞<br />

1 0 1 0<br />

0<br />

D = 〈 ⎜ ⎟<br />

⎝1⎠ , 2<br />

⎜ ⎟<br />

⎝1⎠ , 1<br />

⎜ ⎟<br />

⎝1⎠ , 0<br />

⎜ ⎟<br />

⎝0⎠ 〉<br />

0 0 0 1<br />

a change of basis operation produces this representation with respect to D, D.<br />

⎛<br />

⎞ ⎛<br />

⎞ ⎛<br />

⎞−1<br />

⎛<br />

⎞<br />

1 0 1 0 0 0 0 0 1 0 1 0 −1 0 1 0<br />

0 2 1 0<br />

1 0 0 0<br />

0 2 1 0<br />

−3 −2 5 0<br />

⎜<br />

⎟ ⎜<br />

⎟ ⎜<br />

⎟ = ⎜<br />

⎟<br />

⎝1 1 1 0⎠<br />

⎝0 1 0 0⎠<br />

⎝1 1 1 0⎠<br />

⎝−2 −1 3 0⎠<br />

0 0 0 1 0 0 1 0 0 0 0 1 2 1 −2 0<br />

The new matrix is nilpotent; its fourth power is the zero matrix. We could<br />

verify this with a tedious computation or we can instead just observe that it is<br />

nilpotent since its fourth power is similar to ˆN 4 , the zero matrix, and the only<br />

matrix similar to the zero matrix is itself.<br />

(P ˆNP −1 ) 4 = P ˆNP −1 · P ˆNP −1 · P ˆNP −1 · P ˆNP −1 = P ˆN 4 P −1<br />

The goal of this subsection is to show that the prior example is prototypical<br />

in that every nilpotent matrix is similar to one that is all zeros except for blocks<br />

of subdiagonal ones.<br />

2.11 Definition Let t be a nilpotent transformation on V. At-string generated<br />

by ⃗v ∈ V is a sequence 〈⃗v, t(⃗v),...,t k−1 (⃗v)〉 such that t k (⃗v) =⃗0. A t-string<br />

basis is a basis that is a concatenation of t-strings.<br />

(The strings cannot form a basis under concatenation unless they are disjoint<br />

because a basis cannot have a repeated vector.)<br />

2.12 Example This linear map t: C 3 → C 3<br />

⎛ ⎞ ⎛ ⎞<br />

x<br />

⎜ ⎟<br />

⎝y⎠<br />

z<br />

t<br />

↦−→<br />

y<br />

⎜ ⎟<br />

⎝z⎠<br />

0

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