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Exercise 7.2.14

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Since the characteristic polynomial of $T$ splits and contains the unique zero to be $0$ , $T$ has the Jordan canonical form $J$ whose diagonal entries are all zero. By Exercise 7.2.12, the matrix $J$ is nilpotent. By Exercise 7.2.11, the linear operator $T$ is also nilpotent.

jephianlin

2011-06-27 00:00

Exercise 7.2.14

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