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Exercise 1.6.11

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If ${u, v}$ is a basis then the dimension of $V$ would be two. So it’s enough to check both ${u + v, au}$ and ${au, bv}$ are linearly independent. Assuming $s (u + v) + tau = (s + ta) u + sv = 0$ we have $s + ta = s = 0$ and hence $s = t = 0$ . Assuming $sau + tbv = 0$ we have $sa = tb = 0$ and hence $s = t = 0$ .

jephianlin

2011-06-27 00:00

Exercise 1.6.11

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