answer in steps Let V = {v R4 | Bv = 2v} where B is a particular 4 times 4 matri
ID: 3100351 • Letter: A
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answer in steps
Let V = {v R4 | Bv = 2v} where B is a particular 4 times 4 matrix. This means that V is the set of all vectors in R4 such that Bv = 2v. Either prove that V is a vector space or show (with a counter-example) which condition fails. Let W = {(x, y, z) R3 | x = |y|}. Either prove that W is a vector space or show (with a counter-example) which condition fails.Explanation / Answer
given that B is a fixed matrix such that Bv = 2v suppose u, v are in V then B(u+v) = 2(u+v) = 2u+2v = Bu+Bv and so, u+v is in V so, addition of vectors obey closure law. suppose 0 column is the nx1 matrix, and such that B*0 = 2*0 and so, 0 column is in V. so, 0 column we are referring is the zero vector in V suppose v is in V then Bv = 2v also, we follow that 2(-v) = B(-v) so, -v is in V whenever v is in V that approves the additive inverse vectors in V associativity can simply be established B(u+v) = 2(y+v)=2(v+u) = B(v+u) so, u+v = v+u holds in V a (Bv) = a(2v) = 2(av)=B(av) so, scalar multiplication obeys closure law. (a+b)Bv = (a+b)2v= 2(av+bv) =B(av+bv) so, scalar addition is distributive over addition of vectors B{a(u+v)}=2{a(u+v)}= a(2u+2v) = 2(au)+ 2(av) = B(au)+B(av) ==> a(u+v)= au+av scalar multiplication is distributive over addition of vectors. B(ab(u)) = 2(ab(u)) = a(2(bu))=a(B(bu)) this approves the scalar multiplicative is associativity. B(1v) = 2(1v) = 2v = Bv = 1(Bv) so, all the vector space properties are verified. the given set of vectors in V form a vector space. --------------------------------------------------------------------------- x=|y| leads to x = -y when x < 0 and x = y when x > 0 this says that on negative real numbers, the additive inverse vector of (x,y,z) is different from the additive inverse vector on the positive real numbers. in other words, (x,y,z) has two inverse vectors in R^3. this is a failure property of algebras and not only to vector space structure.Related Questions
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