Then we claim \u03b1T \u03b2U A B \u03b1 T A B \u03b2 U A B Indeed we only need to check this

# Then we claim αt βu a b α t a b β u a b indeed we

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Then we claim [ αT + βU ] A B = α [ T ] A B + β [ U ] A B . Indeed, we only need to check this formula on the basis vectors v 1 , . . . , v n in B . We must have numbers a ij , b ij R such that T ( v j ) = m X i =1 a ij w i , U ( v j ) = m X i =1 b ij w i . 51 Now, ( αT + βU )( v j ) = m X i =1 ( αa ij + βb ij ) w i , which shows [ αT + βU ] A B = α [ a ij ] + β [ b ij ] = α [ T ] A B + β [ U ] A B . 7.2. Dual spaces. Given two vector spaces V and W , we have just seen the vector space L ( V, W ), the vector space of linear transformations from V to W . In the special case that the target W = R , we obtain something called the dual space to V . Definition 15. Let V be a vector space. The dual space V * to V is the vector space V * = L ( V, R ) . An element φ V * is a linear function φ : V R ; such a linear map is often called a linear functional . In the special case that B = { v 1 , . . . , v n } is a basis for V , we have a particular set of n linear functionals B * = { φ 1 , . . . , φ n } , which are defined by the relation φ i ( v j ) = 1 i = j 0 i 6 = j. The main content of the proof of the theorem below is to show that B * is a basis of V * ; it is called the dual basis to B . Theorem 39. Let V be a finite dimensional vector space. Then V * is isomorphic to V . Proof. We pick a basis B = { v 1 , . . . , v n } and prove the theorem. In fact, we al- ready know that V is isomorphic to R n , so we only need to prove that V * is also isomorphic to R n as well. To show this, we only need to show that V * is an n - dimensional vector space, i.e. that V * has a basis with n elements. We already have an excellent candiate for this basis, namely the dual basis B * = { φ 1 , . . . , φ n } . Let φ V * . We show that there is a unique choice of coefficients a 1 , . . . , a n such that φ = a 1 φ 1 + a 2 φ 2 + · · · + a n φ n ; as we have seen before, this shows both that B * spans V * (by the existence of a solution) and that it is linearly independent (by the uniqueness of the solution). Given our choice of φ V * , we can pick a 1 = φ ( v 1 ) , a 2 = φ ( v 2 ) , . . . , a n = φ ( v n ) . Now we compare φ to the map ˜ φ = a 1 φ 1 + a 2 φ 2 + · · · + a n φ n . We have, by definition, φ ( v j ) = a j = ˜ φ ( v j ), which implies φ = ˜ φ . Thus we have represented our arbitrary φ V * as a linear combination of elements of B * . Furthermore, if there is some other choice of coefficients b 1 , . . . , b n such that φ = b 1 φ 1 + · · · b n φ n , 52 then for any j = 1 , . . . , n we have b j = ( b 1 φ 1 + · · · + b n φ n )( v j ) = φ ( v j ) = ( a 1 φ 1 + · · · + a n φ n )( v j ) = a j . Thus the choice of coeffients is unique, completing the proof. Remark 9. Notice that isomorphism V V * depends quite heavily on the choice of basis B (or, equivalently, on the choice of dual basis B * ). If we change basis in V , we will obtain a very different isomorphism V V * . Also notice that in order to construct the dual basis B * , we need to first pick the basis B . We cannot pick them simultaneously, say choosing the first two elements of B , then the elements of B * , then the rest of B . This is because in order to describe even the first linear functional φ 1 ∈ B * , we need to know what it does to each and every basis vector in B .  #### You've reached the end of your free preview.

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• Fall '08
• GUREVITCH
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