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Exercise 8.B.2

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Proof. Define $T \in L (ℝ^{3})$ by

T (x, y, z) = (0, - z, x) .

That is, $T$ squashes vectors onto the $yz$ plane and rotates them counterclockwise by $π ∕ 2$ radians. So all eigenvectors of $T$ are contained in the $x$ -axis and correspond to the eigenvalue $0$ . $T$ obviously is not nilpotent. □

celiopassos

2017-10-06 00:00

Exercise 8.B.2

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