Show that A middot B 2 A times B 2 AB 2 nbsp (2024)

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Show that (A · B)2 + (A × B)2 = (AB)2.

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#### Solution By Steps***Step 1: Expand (A · B)2***(A · B)2 = (A · B) * (A · B) (A · B)2 = A · A · B · B***Step 2: Simplify A · A and B · B***A · A = A^2 (since any vector dotted with itself is the square of its magnitude)B · B = B^2***Step 3: Substitute A^2 and B^2 back into the expression***(A · B)2 = A^2B^2***Step 4: Expand (A × B)2***(A × B)2 = (A × B) * (A × B)(A × B)2 = A × A × B × B***Step 5: Simplify A × A and B × B***A × A = 0 (since the cross product of a vector with itself is the zero vector)B × B = 0***Step 6: Substitute 0 back into the expression***(A × B)2 = 0***Step 7: Add the results of (A · B)2 and (A × B)2***(A · B)2 + (A × B)2 = A^2B^2 + 0***Step 8: Simplify the expression***(A · B)2 + (A × B)2 = A^2B^2#### Final Answer(A · B)2 + (A × B)2 = (AB)2

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An algebra is a vector space A over K together with a multiplication operator middot AtimesA rarr Asuch thatbull a middot (b middot c) = (a middot b) middot c for all a b c isin Abull a middot (b + c) = (a middot b) + (a middot c) and (a + b) middot c = (a middot c) + (b middot c) for all a b c isin A andbull (lambdaa) middot b = lambda(a middot b) = a middot (lambdab) for all a b isin A and lambda isin KIf there exists an element I isin A such that Ia = aI = a for all a isin A we call I the unit of A and say A isunital If A has a unit it is elementary to verify that A has a unique unit by algebraic argumentsA normed linear space (A middot ) is said to be a unital normed algebra if A is a unital algebra I = 1and ab le a b for all a b isin A For example given any normed linear space (X middot X) we have thatB(X ) together with the operator norm is a unital normed algebra where the unit is the identity mapProve that if (A middot ) is a normed algebra and (an)nge1 and (bn)nge1 are sequences in A that convergeto aprime and bprime respectively then (anbn)nge1 is a convergent sequence that converges to aprimebprimeAnswered step-by-step 1 answerAssume that B is a Boolean algebra with operations + and middot Prove the following statement De Morgan39s law for + For all a and b in B a + b = a middot b Proof Suppose B is a Boolean algebra and a and b are any elements of B We must show that a + b = a middot b Part 1 Proof that (a + b) middot (a middot b) = 0 (a + b) middot (a middot b) = (a middot b) middot (a + b) by the commutative law for middot = (a middot b) middot a + (a middot b) middot b by the distributive law of middot over + = (b middot a) middot a + (a middot b) middot b by the commutative law for middot = b middot (a middot a) + a middot (b middot b) by the associative law for middot = b middot (a middot a) + a middot (b middot b) by the commutative law for middot = (b middot 0) + (a middot 0) by the complement law for middot = 0 + 0 by the universal bound law for middot = 0Answered step-by-step 1 answerProve that (AB)2=A2 + AB + BA + B2Answered step-by-step 1 answerProve that (a2+1ampabampacabampb2+1ampbcacampbcampc2+1) =1 + a 2 + b 2 + c 2Answered step-by-step 1 answerShow that (A middot B)2 + (A times B)2 = (AB)2Answered step-by-step 2 answers

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Show that A middot B 2 A times B 2 AB 2 nbsp (2024)

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