- #1

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what is the general way of solving this question:

http://img116.imageshack.us/img116/1152/25587465vv2.gif [Broken]

??

http://img116.imageshack.us/img116/1152/25587465vv2.gif [Broken]

??

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- Thread starter transgalactic
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- #1

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http://img116.imageshack.us/img116/1152/25587465vv2.gif [Broken]

??

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- #2

Defennder

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2. Firstly determine the solution space of U(k). Then [tex]span(U(k1)) \subseteq span(V(k2)) [/tex] if [tex]\forall v \ \text{where v is a vector in basis of U(k1)} \ , v \in span(V(k2))[/tex]. ie. try to express every vector in the basis of U(k1) as a linear combination of V(k2). Take note of when this is possible (ie. which values of k1 and k2 permit that).

- #3

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i think i solved part 1.

but i dont know how to use part 1

in order to solve 2

http://img384.imageshack.us/img384/2546/55339538nk4.gif [Broken]

but i dont know how to use part 1

in order to solve 2

http://img384.imageshack.us/img384/2546/55339538nk4.gif [Broken]

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- #4

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i can find k values for which i get one solution

infinite solution

or no solution

what is the definition of solution space in a parameter matrix?

- #5

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how do i find the solution vectors of U(k)?

- #6

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"Firstly determine the solution space of U(k). "

there are 3 types of solution?(depends on the values of K)

no solution

1 solution

infinite solution

- #7

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