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Exercise 2.3.15

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Let $v_{t}$ be the $t$ -th column vector of A and we have $v_{j} = \sum_{t \neq j} a_{t} v_{t}$ . Thus we have $M v_{j} = \sum_{t \neq j} a_{t} M v_{t}$ . And hence we get the desired result since $M v_{t}$ is the column vector of $MA$ .

jephianlin

2011-06-27 00:00

Exercise 2.3.15

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