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Abstract Algebra Theory and Applications - Computer Science ...

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10.1 MATRIX GROUPS 175We can define the length of a vector using the Euclidean inner product,or dot product, of two vectors. The Euclidean inner product of two vectorsx = (x 1 , . . . , x n ) t <strong>and</strong> y = (y 1 , . . . , y n ) t is⎛〈x, y〉 = x t y = (x 1 , x 2 , . . . , x n ) ⎜⎝⎞y 1y 2⎟.y nWe define the length of a vector x = (x 1 , . . . , x n ) t to be‖x‖ = √ √〈x, x〉 = x 2 1 + · · · + x2 n.⎠ = x 1y 1 + · · · + x n y n .Associated with the notion of the length of a vector is the idea of the distancebetween two vectors. We define the distance between two vectors x <strong>and</strong> yto be ‖x − y‖. We leave as an exercise the proof of the following propositionabout the properties of Euclidean inner products.Proposition 10.1 Let x, y, <strong>and</strong> w be vectors in R n <strong>and</strong> α ∈ R. Then1. 〈x, y〉 = 〈y, x〉.2. 〈x, y + w〉 = 〈x, y〉 + 〈x, w〉.3. 〈αx, y〉 = 〈x, αy〉 = α〈x, y〉.4. 〈x, x〉 ≥ 0 with equality exactly when x = 0.5. If 〈x, y〉 = 0 for all x in R n , then y = 0.Example 5. The vector x = (3, 4) t has length √ 3 2 + 4 2 = 5. We can alsosee that the orthogonal matrix( )3/5 −4/5A =4/5 3/5preserves the length of this vector. The vector Ax = (−7/5, 24/5) t also haslength 5.Since det(AA t ) = det(I) = 1 <strong>and</strong> det(A) = det(A t ), the determinant ofany orthogonal matrix is either 1 or −1. Consider the column vectors⎛ ⎞a 1ja 2ja j = ⎜ ⎟⎝ . ⎠a nj

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