Exercise 8.A.1

Answers

Proof. Since

T^{2} (w, z) = T (z, 0) = (0, 0),

it follows that $G (0, T) = V$ . Therefore every vector in $ℂ^{2}$ is a generalized eigenvector of $T$ . □

celiopassos

2017-10-06 00:00