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Material Type: Quiz; Class: Applied Linear Algebra; Subject: Mathematics; University: University of Illinois - Chicago; Term: Unknown 2012;
Typology: Quizzes
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Self-quiz 1
E^3 = {(a, b, c) : a, b, c real numbers}
We define addition in E^3 by:
(a 1 , b 1 , c 1 ) + (a 2 , b 2 , c 2 ) = (a 1 + a 2 , b 1 + b 2 , c 1 + c 2 )
and scalar multiplication by:
λ(a, b, c) = (λa, λb, λc)
Show that E^3 equipped with these two operations is a vector space.
3 v + 2u + w = 0
where u = (1, −1)T^ and w = (2, 1)T^.
This implies that:
((a 1 , b 1 , c 1 )+(a 2 , b 2 , c 2 ))+(a 3 , b 3 , c 3 ) = (a 1 , b 1 , c 1 )+((a 2 , b 2 , c 2 )+(a 3 , b 3 , c 3 ))
i.e. addition is indeed associative.
(a, b, c) + (0, 0 , 0) = (a, b, c)
(a, b, c) + (−a, −b, −c) = (0, 0 , 0)
(λ + μ)(a, b, c) = λ(a, b, c) + μ(a, b, c)
Indeed each side equals (λa + μa, λb + μb, λc + μc) as can be easily seen by carrying out the computation
(λa 1 + λa 2 , λb 1 + λb 2 , λc 1 + λc 2 )
(λμa, λμb, λμc)
So E^3 is indeed a vector space when equipped with the above operations.
3 v + 2u + w = 0
where u = (1, −1)T^ and w = (2, 1)T^.
Solution: We can solve the given equation for v as follows:
3 v = − 2 u − w =⇒ v = −
3 u^ −^
3 w
By substituting u and w, we get:
v = −
