Exercise 9.A.14

Answers

Proof. Because it is nilpotent, the minimal polynomial of $T_{}$

C^2 + T_C + I $i s o f t h e f o r m$ z^m $f o r s o m e p o s i t i v e i n t e g e r$ m $. W e h a v e$

z^{2} + z + 1 = (z - λ) (z - \bar{λ})

for some nonreal $λ \in ℂ$ . Thus

0 = (T_{}

C^2 + T_C + I)^m = (T_C - $λ)^{m} (T_{}$

C - $λ$ )^m.

This, together with 9.16, implies that the eigenvalues of $T_{}$

C $a r e$ $λ$ and $\bar{λ}$ , with equal multiplicities, namely $4$ . The characteristic polynomial of $T_{}$

C $, a n d o f$ T $a s w e l l b y d e f i n i t i o n, i s t h e r e f o r e$

{(z - λ)}^{4} {(z - \bar{λ})}^{4} .

which equals ${(z^{2} + z + 1)}^{4}$ . By the Cayley-Hamilton Theorem (9.24), it follows that

{(T^{2} + T + I)}^{4} = 0 .

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celiopassos

2017-10-06 00:00