Exercise 2.3.5

Answers

We have
$\begin{array}{l} {(a (AB))}_{ij} & = a \sum_{k = 1}^{n} A_{ik} B_{kj} \\ \sum_{k = 1}^{n} a A_{ik} B_{kj} = {((aA) B)}_{ij} \\ \sum_{k = 1}^{n} A_{ik} a B_{kj} = {(A (aB))}_{ij} \end{array}$
We have ${[I (v_{i})]}_{α} = e_{i}$ where $v_{i}$ is the $i$ -th vector of $β$ .

We have by Theorem 2.12

A (\sum_{i = 1}^{k} a_{i} B_{i}) = \sum_{i = 1}^{k} A (a_{i} B_{i}) = \sum_{i = 1}^{k} a_{i} A B_{i}

and

(\sum_{i = 1}^{k} a_{i} C_{i}) A = \sum_{i = 1}^{k} (a_{i} C_{i}) A = \sum_{i = 1}^{k} a_{i} C_{i} A .

jephianlin

2011-06-27 00:00