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Exercise 2.1.3

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Similarly check this is a linear transformation. $N (T) = {0}$ with basis $\emptyset$ ; $R (T) = {a_{1} (1, 0, 2) + a_{2} (1, 0, - 1)}$ with basis ${(1, 0, 2), (1, 0, - 1)}$ . Hence $T$ is one-to-one but not onto.

jephianlin

2011-06-27 00:00

Exercise 2.1.3

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