Problem: (1998 IMC Day 1 - #1) Let $ V $ be a $ 10 $-dimensional real vector space and $ U_1, U_2 $ two linear subspaces such that $ U_1 \subseteq U_2 $, $ \dim U_1 = 3 $, and $ \dim U_2 = 6 $. Let $ \epsilon $ be the set of linear maps $ T: V \rightarrow V $ which have $ T(U_1) \subseteq U_1 $ and $ T(U_2) \subseteq U_2 $. Calculate the dimension of $ \epsilon $.
Solution: Well we have a total of $ 10 \cdot 10 = 100 $ degrees of freedom without the constraints. The constraint $ T(U_1) \subseteq U_1 $ takes away $ 3 \cdot 7 = 21 $ (since all three dimensions of $ U_1 $ have to map to $ U_1 $), while the constraint $ T(U_2) \subseteq U_2 $ takes away an additional $ 3 \cdot 4 = 12 $ (since the remaining three dimensions of $ U_2 $ have to map to $ U_2 $). Hence the total number of dimensions of linear maps is $ 100-21-12 = 67 $. QED.
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Comment: A pretty standard problem on linear transformations and linear algebra. Easy by most standards, since it really only involves some simple calculations.
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Practice Problem: What is the dimension of the span of the vectors $ (1, 2, 3) $, $ (2, 3, 4) $, and $ (0, -1, -2) $?
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