Explanation Given: The matrices, A = ( a i j ) and B = ( b i j ) are of order ( 2 × 2 ) And 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 Approach: An inner product on a real vector space V is a function that assigns a real number, 〈 u , v 〉 , to each pair of vectors u and v , and that satisfies the following properties: 1) 〈 u , u 〉 ≥ 0 and 〈 u , u 〉 = 0 if and only if u = 0 . 2) 〈 u , v 〉 = 〈 v , u 〉 3) 〈 a u , v 〉 = a 〈 u , v 〉 4) 〈 u , v + w 〉 = 〈 u , v 〉 + 〈 u , w 〉 Calculation: Property (1) For the vector space of all the ( 2 × 2 ) matrices, the expression is 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 So, by Property (1) 〈 A , A 〉 = a 11 a 11 + a 12 a 12 + a 21 a 21 + a 22 a 22 = ( a 11 ) 2 + ( a 12 ) 2 + ( a 21 ) 2 + ( a 22 ) 2 Since, the square of a real number is always positive, this implies that [ ( a 11 ) 2 + ( a 12 ) 2 + ( a 21 ) 2 + ( a 22 ) 2 ] > 0 So, 〈 A , A 〉 > 0 Also, 〈 A , A 〉 = 0 if and only if all of the a 11 , a 12 , a 21 , a 22 are zero that is, A = 0 . Thus, 〈 A , A 〉 ≥ 0 Hence, property (1) is satisfied. (Property-2) For the vector space of all the ( 2 × 2 ) matrices, the expression is 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 So, by the property (2) 〈 B , A 〉 = b 11 a 11 + b 12 a 12 + b 21 a 21 + b 22 a 22 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 = 〈 A , B 〉 Hence, property (2) is satisfied. (Property-3) For the vector space of all the ( 2 × 2 ) matrices, the given expression is 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 Assume c be any real scalar, then c A = c [ a 11 a 12 a 21 a 22 ] = [ c a 11 c a 12 c a 21 c a 22 ] And, B = [ b 11 b 12 b 21 b 22 ] So, by property (3)

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

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ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 5.6, Problem 7E

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The expression 〈A,B〉=a11b11+a12b12+a21b21+a22b22 is an inner product for the vector space of all (2×2) matrices.

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Chapter 5.6, Problem 6E

Chapter 5.6, Problem 8E

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Q1lal Let X be an arbitrary infinite set and let r the family of all subsets F of X which do not contain a particular point x, EX and the complements F of all finite subsets F of X show that (X.r) is a topology. bl The nbhd system N(x) at x in a topological space X has the following properties NO- N(x) for any xX N1- If N EN(x) then x€N N2- If NEN(x), NCM then MeN(x) N3- If NEN(x), MEN(x) then NOMEN(x) N4- If N = N(x) then 3M = N(x) such that MCN then MeN(y) for any уем Show that there exist a unique topology τ on X. Q2\a\let (X,r) be the topology space and BST show that ẞ is base for a topology on X iff for any G open set xEG then there exist A Eẞ such that x E ACG. b\Let ẞ is a collection of open sets in X show that is base for a topology on X iff for each xex the collection B, (BEB\xEB) is is a nbhd base at x. - Q31 Choose only two: al Let A be a subspace of a space X show that FCA is closed iff F KOA, K is closed set in X. الرياضيات b\ Let X and Y be two topological space and f:X -…

Q1\ Let X be a topological space and let Int be the interior operation defined on P(X) such that 1₁.Int(X) = X 12. Int (A) CA for each A = P(X) 13. Int (int (A) = Int (A) for each A = P(X) 14. Int (An B) = Int(A) n Int (B) for each A, B = P(X) 15. A is open iff Int (A) = A Show that there exist a unique topology T on X. Q2\ Let X be a topological space and suppose that a nbhd base has been fixed at each x E X and A SCX show that A open iff A contains a basic nbdh of each its point Q3\ Let X be a topological space and and A CX show that A closed set iff every limit point of A is in A. A'S A ACA Q4\ If ẞ is a collection of open sets in X show that ẞ is a base for a topology on X iff for each x E X then ẞx = {BE B|x E B} is a nbhd base at x. Q5\ If A subspace of a topological space X, if x Є A show that V is nbhd of x in A iff V = Un A where U is nbdh of x in X.

Chapter 5 Solutions

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - Prob. 7E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...

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The expression 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 is an inner product for the vector space of all ( 2 × 2 ) matrices.

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The expression 〈A,B〉=a11b11+a12b12+a21b21+a22b22 is an inner product for the vector space of all (2×2) matrices.

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Chapter 5 Solutions

The expression 〈 A , B 〉 = a 11 b 11 + a 12 b 12 + a 21 b 21 + a 22 b 22 is an inner product for the vector space of all ( 2 × 2 ) matrices.

Solution Summary: The author explains that the expression langle A,Brangle =a_11b­1+12i’j) is an inner product

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To show: The expression 〈A,B〉=a11b11+a12b12+a21b21+a22b22 is an inner product for the vector space of all (2×2) matrices.

Want to see the full answer?

Chapter 5 Solutions

Solution Summary: The author explains that the expression langle A,Brangle =a_11b1+12i’j) is an inner product

To show:

The expression 〈A,B〉=a11b11+a12b12+a21b21+a22b22 is an inner product for the vector space of all (2×2) matrices.