Exercise 7.1.1

Answers

1.
Yes. It comes directly from the definition.
2.
No. If x is a generalized eigenvector, we can find the smallest positive integer p such that (T − λI)p(x) = 0. Thus y = (T − λI)p−1≠0 is an eigenvector with respect to the eigenvalue λ. Hence λ must be an eigenvalue.
3.
No. To apply the theorems in this section, the characteristic polynomial should split. For example, the matrix (0−1 1 0 ) over ℝ has no eigenvalues.
4.
Yes. This is a result of Theorem 7.6.
5.
No. The identity mapping I2 from ℂ2 to ℂ2 has two cycles for the eigenvalue 1.
6.
No. The basis βi may not consisting of a union of cycles. For example, the transformation T(a,b) = (a + b,b) has only one eigenvalue 1. The generalized eigenspace K1 = 𝔽2. If
β = {(1,1),(1,−1)},

then the matrix representation would be

[T]β = 1 2 (3−1 1 1 ),

which is not a Jordan form.

7.
Yes. Let α be the standard basis. Then [LJ]α = J is a Jordan form.
8.
Yes. This is Theorem 7.2.
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2011-06-27 00:00
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