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Exercise 6.5.10

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Exercise 2.5.10 says that $tr (A) = tr (B)$ is $A$ is similar to $B$ . And we know that $A$ may be diagonalized as $P^{∗} AP = D$ by Theorem 6.19 and Theorem 6.20. Here $D$ is a diagonal matrix whose diagonal entries consist of all eigenvalues. This means

tr (A) = tr (D) = \sum_{i = 1}^{n} λ_{i}

and

\begin{array}{l} tr (A^{∗} A) & = tr ({(PD P^{∗})}^{∗} (PD P^{∗})) \\ = tr (P D^{∗} D P^{∗}) = tr (D^{∗} D) = \sum_{i = 1}^{n} | λ_{i} |^{2} . \end{array}

jephianlin

2011-06-27 00:00

Exercise 6.5.10

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