I. N. Herstein's Abstract Algebra (3rd Edition) PDF

By I. N. Herstein

ISBN-10: 0133745627

ISBN-13: 9780133745627

Starting summary Algebra with the vintage Herstein remedy.

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Extra resources for Abstract Algebra (3rd Edition)

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The operation is 4R2 + 6R3 → R3 , and so we replace the element at the second column with a 4, and the element in the third column with a 6, which results in the matrix   1 0 0 0 1 0 0 4 6 EXAMPLE 1-10 For a 4 × 4 matrix, find the elementary matrix that represents −2R2 + 5R4 → R4 21 22 CHAPTER 1 Systems of Linear Equations SOLUTION 1-10 To construct an elementary matrix, we begin with a matrix with 1s along the diagonal and 0s everywhere else. For a 4 × 4 matrix, we use  1 0 I4 =  0 0 0 1 0 0 0 0 1 0  0 0 0 1 The destination row is the fourth row, and so we will modify the fourth row of I4 .

We summarize this by writing AT + B T ij = aji + bji On the left-hand side, the (i, j) entry of C = A + B is cij = aij + bij The (i, j) element of C T is also found by swapping row and column indices, and so the (i, j) entry of (A + B)T is (A + B)ijT = aji + bji The (i, j) element of both sides are equal; therefore, the matrices are the same. The Hermitian Conjugate We now extend the transpose operation to the Hermitian conjugate, which is written as A† (read “A dagger”). The Hermitian conjugate applies to matrices with complex elements and is a two-step operation (see Fig.

Matrix A Step 1 Transpose Step 2 Conjugate ∗ Fig. 2-4. A schematic representation of finding the Hermitian conjugate of a matrix. Take the transpose, turning rows into columns, and then compute the complex conjugate of each element. CHAPTER 2 50 EXAMPLE 2-10 Find A† for  −9  0 A= 1 + 2i 2i 4i 3 Matrix Algebra  0 7 0 SOLUTION 2-10 Step 1, we take the transpose of the matrix:  T  −9 2i 0 −9 0 T    0 4i 7 A = = 2i 4i 1 + 2i 3 0 0 7  1 + 2i 3  0 Now we apply step 2 by taking the complex conjugate of each element.

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Abstract Algebra (3rd Edition) by I. N. Herstein


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