## Thursday, November 9, 2006

### Dimensions Everywhere. Topic: Linear Algebra.

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 $(0, -1, -2)$?