Download Vector Calculus, Linear Algebra and Differential Forms: A by John Hamal Hubbard, Barbara Burke Hubbard PDF

By John Hamal Hubbard, Barbara Burke Hubbard

Utilizing a dual-presentation that's rigorous and comprehensive--yet tremendously "student-friendly" in approach--this textual content covers lots of the usual subject matters in multivariate calculus and a considerable a part of a regular first path in linear algebra. It makes a speciality of underlying rules, integrates thought and functions, bargains a number of pedagogical aids, and contours assurance of differential kinds. there's an emphasis on numerical easy methods to arrange scholars for contemporary purposes of arithmetic.

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Extra resources for Vector Calculus, Linear Algebra and Differential Forms: A Unified Approach

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That n = q = 1, so that both (AB)C and A(BC) are numbers. 5. (AB)C = > (> akbk,i J ct 1=1 This way of writing the matrices corresponds to calculating A(BC). k=l / tth entry of AB =L 1=1k=1 akbk,lc! 9 A(BC). 40 Chapter 1. 2 provides practice on matrix multiplication. 22, involving graphs, shows a setting where matrix multiplication is a natural and powerful tool. As we saw earlier, matrix multiplication is most definitely not commutative. It may well be possible to multiply A by B but not B by A.

Schroeder, "Number Theory and the Real World," Mathematical lntelligencer, Vol. 7, No. 1 Given the matrices A- [2 30] B= [o 11 C = [1 -10 -1i D= 2 [11 2 1 01 what are the products All, AC and CD? 2 Now compute BA. What do you notice? 3 2AB=[0 5]; AC=[b -2 _i] ; CD= g 3Matrix multiplication is not commutative; BA = _ij. 3}, which is not equal to 12 2 AB = 1 0 5l . Although the product AC exists, you cannot compute CA. 2 Matrices :17 Below we state the formal definition of the process we've just, ch'scribed.

5 (Subspace of IR"). , V is a subspace if when x, y E V , and a E 118, To he closed under multiplication a subspace must contain the zero vector, so that then z+ y E V and arZ E V. For example, a straight line through the origin is a subspace of II22 and of 1F3. A plane through the origin is a subspace of 1183. The set consisting of just the zero vector {ti} is a subspace of any R1, and H8" is a subspace of itself. These last two, {0} and 18", are considered trivial subspaces. Intuitively, it is clear that a line that is a subspace has dimension 1, and a plane that is a subspace has dimension 2.

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