By Gallier J.

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Un ) to (v1 , . . , vn ), so that n vj = a i j ui , i=1 and let P −1 = (bi j ) be the inverse of P , so that n bj i v j . ui = j=1 Since u∗i (uj ) = δi j and vi∗ (vj ) = δi j , we get n vj∗ (ui ) = vj∗ ( bk i v k ) = b j i , k=1 and thus n vj∗ bj i u∗i , = i=1 and similarly n u∗i ai j vj∗ . = j=1 Since n ϕ∗ = n ϕi u∗i = i=1 we get ϕi vi∗ , i=1 n ai j ϕ i . ϕj = i=1 Comparing with the change of basis n vj = a i j ui , i=1 48 CHAPTER 2. LINEAR ALGEBRA we note that this time, the coordinates (ϕi ) of the linear form ϕ change in the same direction as the change of basis.

Vm ). t. the bases (u1 , . . , un ) and (v1 , . . t. the bases (u1 , . . , un ) and (v1 , . . , vm ). We have M (f ) = Q−1 M (f )P. Proof . t. the bases (u1 , . . , un ) and (u1 , . . t. the bases (v1 , . . , vm ) and (v1 , . . 12, we have M (f ) = Q−1 M (f )P . As a corollary, we get the following result. 36 CHAPTER 2. 15 Let E be a vector space, and let (u1 , . . , un ) and (u1 , . . , un ) be two bases of E. Let P be the change of basis matrix from (u1 , . . , un ) to (u1 , .

Xi , xi , . ) + f (. . , xi , xi+1 , . ) +f (. . , xi+1 , xi , . ) + f (. . , xi+1 , xi+1 , . ), and since f is alternating, this yields 0 = f (. . , xi , xi+1 , . ) + f (. . , xi+1 , xi , . ), that is, f (. . , xi , xi+1 , . ) = −f (. . , xi+1 , xi , . ). (2) If xi = xj and i and j are not adjacent, we can interchange xi and xi+1 , and then xi and xi+2 , etc, until xi and xj become adjacent. By (1), f (. . , xi , . . , xj , . ) = f (. . , xi , xj , . ), where = +1 or −1, but f (.