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Exercise 9.A.19

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Proof. By the same reasoning used in the proof 8.4, it follows that $\dim null T^{n - 1} \geq n - 1$ . Thus, the multiplicty of $0$ as an eigenvalue of $T$ , and $T_{}$

C $a s w e l l, i s a t l e a s t$ n - 1 $. H e n c e$ T_C $h a s a t m o s t a n o t h e r e i g e n v a l u e . S i n c e n o n r e a l e i g e n v a l u e s o f$ T_C $c o m e i n p a i r s (s e e 9.16), t h i s o t h e r e i g e n v a l u e o f$ T_C $m u s t b e r e a l . □$

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2017-10-06 00:00

Exercise 9.A.19

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