Exercise 2.3.13

Answers

It’s natural that we have tr $(A) =$ tr $(A^{t})$ since $A_{ii} = A_{ii}^{t}$ for all $i$ . On the other hand we have

\begin{array}{l} tr (AB) & = \sum_{i = 1}^{n} {(AB)}_{ii} = \sum_{i = 1}^{n} \sum_{k = 1}^{n} A_{ik} B_{ki} \\ = \sum_{k = 1}^{n} \sum_{ik = 1}^{n} B_{ki} A_{ik} = \sum_{k = 1}^{n} {(BA)}_{kk} \\ = tr (BA) . \end{array}

jephianlin

2011-06-27 00:00