By Cao H. D., Chow B., Chu S. C., Yau S. T. (eds.)

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**Extra resources for Collected papers on Ricci flow**

**Example text**

21. Let (V, G) be a finite-dimensional inner product space, and let T : V → V be a linear isometry. Then its inverse T −1 is also a linear isometry. Proof. 16 to G = (Id)∗ G = (T ◦ T −1 )∗ G and use the assumption that T ∗ G = G. We conclude this section with an important technical theorem, a consequence of the positive definite property of inner products. 5, the two vector spaces are isomorphic. A choice of an inner product on V , however, induces a distinguished isomorphism between them. 22. Let G be an inner product on a finite-dimensional vector space V .

Xk−1 , y), and T (sx1 , x2 , . . , xk ) = sT (x1 , x2 , . . , xk ), .. T (x1 , x2 , . . , sxk ) = sT (x1 , x2 , . . , xk ). In the special case that all the Vi are the same and W = R, then a multilinear function T : V × · · · × V → R is called a multilinear k-form on V . 9 (The zero k-form on V ). The trivial example of a k-form on a vector space V is the zero form. Define O(v1 , . . , vk ) = 0 for all v1 , . . , vk ∈ V . We leave it to the reader to show that O is multilinear. 10 (The determinant as an n-form on Rn ).

Note that for a general vector w = (w1 , w2 ) ∈ R2 , writing w as a linear combination of B , w = c1 e1 + c2 e2 amounts to solving the system c1 (−1) + c2 (2) = w1 , c1 (1) + c2 (1) = w2 . 3. However, to illustrate an efficient general method for finding the matrix representation of a linear transformation, let us solve this system simultaneously for T (e1 ) = (2, 1), T (e2 ) = (3, 1), and T (e3 ) = (2, 4) by Gaussian elimination of the matrix 26 2 Linear Algebra Essentials −1 2 1 1 2 1 3 1 0 1 0 1 −1/3 4/3 2 4 , yielding 1 0 2 2 .