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Linear Algebra

Exercise 2.3.6

Answers

We have ${(B e_{j})}_{i} = \sum_{k = 1}^{n} B_{ik} {(e_{j})}_{i} = B_{ij}$ , since ${(e_{j})}_{i} = 1$ only when $i = j$ and it’s $0$ otherwise.

jephianlin

2011-06-27 00:00

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