By Professor Martin Anthony, Dr Michele Harvey
Any pupil of linear algebra will welcome this textbook, which supplies a radical remedy of this key subject. mixing perform and concept, the publication permits the reader to profit and understand the traditional tools, with an emphasis on knowing how they really paintings. At each level, the authors are cautious to make sure that the dialogue is not any extra advanced or summary than it has to be, and makes a speciality of the elemental themes. The booklet is perfect as a direction textual content or for self-study. teachers can draw at the many examples and workouts to complement their very own assignments. End-of-chapter sections summarize the fabric to aid scholars consolidate their studying as they development throughout the publication.
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Additional info for Linear Algebra: Concepts and Methods
To prove this is true, we just need to show that the matrix C satisfies (λA)C = C(λA) = I . This is straightforward using matrix algebra: r (λA)−1 = (λA) 1 1 −1 = λ A A−1 = I and A λ λ 1 1 −1 (λA) = λA−1 A = I. A λ λ If A and B are invertible n × n matrices, then using the definition of the inverse you can show the following important fact: r (AB)−1 = B −1 A−1 . 20 Matrices and vectors This last statement says that if A and B are invertible matrices of the same size, then the product AB is invertible and its inverse is the product of the inverses in the reverse order.
Vk in Rn and scalars α1 , α2 , . . , αk in R, the vector v = α1 v1 + · · · + αk vk ∈ Rn is known as a linear combination of the vectors v1 , . . , vk . A zero vector, denoted 0, is a vector with all of its entries equal to 0. There is one zero vector in each space Rn . As with matrices, this vector is an additive identity, meaning that for any vector v ∈ Rn , 0 + v = v + 0 = v. Further, multiplying any vector v by the scalar zero results in the zero vector: 0v = 0. Although the matrix product of two vectors v and w in Rn cannot be calculated, it is possible to form the matrix products vT w and vwT .
13 If A is m × n and B is n × p, then AB is an m × p matrix. The size of a matrix is not changed by scalar multiplication, so both λ(AB) and (λA)B are m × p. Looking at the (i, j) entries of each, (λ(AB))i j = λ ai1 b1 j + ai2 b2 j + . . + ain bn j = λai1 b1 j + λai2 b2 j + . . + λain bn j = ((λA)B)i j , so these two matrices are equal. 16 In this case, I is m × m. 22 AB = 3 2 1 − 12 1 3 −2 1 . − 12 3 2 2 4 1 − 12 = = 1 0 1 0 0 1 0 . 24 We will show one way (namely, that A A−1 = I ), but you should also show that A−1 A = I .