38 proof we must show that b is a basis for v first

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Proof. We must show that B * is a basis for V * . First, we show B * is linearly independent. Suppose n X i =1 λ i v * i = 0 , the zero linear transformation. Then we have that n X i =1 λ i v * i ! v j = λ j = 0 for all j = 1 , . . . , n . Thus, B * is linearly independent. We show B * spans V * . Suppose ϕ B * . Let λ j = ϕ ( v j ) for j = 1 , . . . , n . Then we have that ϕ - n X i =1 λ i v * i ! v j = 0 for all j = 1 , . . . , n . Since a linear transformation is completely determined by its values on a basis, we have that ϕ - n X i =1 λ i v * i = 0 = ϕ = n X i =1 λ i v * i . Remark 3.3.4 . One of the most useful techniques to show linear independence is the “kill-off” method used in the previous proof: n X i =1 λ i v * i ! v j = λ j = 0 . Applying the v j allows us to see that each λ j is zero. We will see this technique again when we discuss eigenvalues and eigenvectors. Proposition 3.3.5. Let V be a finite dimensional vector space over F . There is a canonical isomorphism Ψ: V V ** = ( V * ) * given by v 7→ ev v , the linear transformation V * F given by the evaluation map: ev v ( ϕ ) = ϕ ( v ) . Proof. First, it is clear that Ψ is a linear transformation as ev λu + v = λ ev u + ev v . Hence, we must show the map Ψ is bijective. Suppose Ψ v = 0. Then ev v is the zero linear functional on V * , i.e., for all ϕ V * , ϕ ( v ) = 0. Since ϕ is completely determined by where it sends a basis of V , we know that ϕ = 0. Hence, the map Ψ is injective. Suppose now that x V ** . Let B = { v 1 , . . . , v n } be a basis for V and let B * = { v * 1 , . . . , v * n } be the dual basis. Setting λ j = x ( v * j ) for j = 1 , . . . , n , we see that if u = n X i =1 λ i v i , 39
then ( x - ev u ) v * j = 0 for all j = 1 , . . . , n . Once more since a linear transformation is completely determined by where it sends a basis, we have x - ev u = 0, so x = ev u and Ψ is surjective. Proposition 3.3.6. Let V be a finite dimensional vector space over F . Then V is isomorphic to V * , but not canonically. Proof. Let B = { v 1 , . . . , v n } be a basis for V . Then B * is a basis for V * . Define a linear transformation T : V V * by v i 7→ v * i for all v i B , and extend this map by linearity. Then T is an isomorphism as there is the obvious inverse map. 3.4 Coordinates For this section, V, W will denote finite dimensional vector spaced over F . We now discuss the way in which all finite dimensional vector spaces over F look like F n (non-uniquely). We then study L ( V, W ), with the main result being that if dim( V ) = n and dim( W ) = m , then L ( V, W ) = M m × n ( F ) (non-uniquely). Definition 3.4.1. Let V be a finite dimensional vector space over F , and let B = { v 1 , . . . , v n } be an ordered basis for V . The map [ · ] B : V R n given by n X i =1 λ i v i 7-→ λ 1 λ 2 . . . λ n is called the coordinate map with respect to B . We say that [ v ] B are coordinates for v with respect to the basis B . Proposition 3.4.2. The coordinate map is an isomorphism. Proof. It is clear that the coordinate map is a linear transformation as [ λu + v ] B = λ [ u ] B +[ v ] B for all u, v V and λ F . We must show [ · ] B is bijective. We show it is injective. Suppose that [ v ] B = (0 , . . . , 0) T . Then since v = n X i =1 λ i v i for unique λ 1 , . . . , λ n F , we must have that λ i = 0 for all i = 1 , . . . , n , and v = 0. Hence the map is injective.

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