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Math 2040 Matrix Theory and Linear Algebra II sample midterm ...

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[ ]<br />

[ ]<br />

r1<br />

x1<br />

(e) (Abstract vector spaces.) Let V = { | r<br />

r 1 +r 2 = 1 <strong>and</strong> r 1 , r 2 ∈ R}. For any ∈ V ,<br />

2 x<br />

[ ]<br />

[ ] [ ]<br />

[ 2<br />

]<br />

y1<br />

x1 y1<br />

x1 + y<br />

∈ V , <strong>and</strong> α ∈ R, suppose + in V is defined to be<br />

1 − 1<br />

<strong>and</strong><br />

y 2 x 2 y 2 x 2 + y<br />

[ ]<br />

[ ]<br />

2<br />

x1<br />

αx1 + (1 − α)<br />

α in V is defined to be<br />

. It turns out that V is a vector space with<br />

x 2 αx 2<br />

this addition <strong>and</strong> scalar multiplication defined on it. Answer the following four questions<br />

about the vector space V .<br />

i. What is 0 ∈ V [<br />

]<br />

1<br />

Solution: 0 = in V .<br />

0<br />

[ ]<br />

4<br />

ii. What is the additive inverse of in V <br />

−3<br />

[ ]<br />

[ ] [ ]<br />

−2 4<br />

1<br />

Solution: , since adding this to results in .<br />

3<br />

−3<br />

0<br />

[ ]<br />

4<br />

iii. What is 5 in V <br />

−3<br />

[ ] [ ]<br />

5(4) + (1 − 5)<br />

16<br />

Solution:<br />

; that is, .<br />

5(−3)<br />

−15<br />

iv. V is a subset of R 2 , there is a 0 ∈ V , <strong>and</strong> the given addition <strong>and</strong> multiplication by<br />

scalars are closed operations on V . However, V a not subspace of R 2 with this vector<br />

space structure defined on [ V . Why ]<br />

0<br />

Solution: The 0 ∈ R 2 is , but this is not in V . Also, the addition <strong>and</strong> scalar<br />

0<br />

multiplication given for V does not correspond to addition <strong>and</strong> scalar multiplication<br />

inherited from R 2 .<br />

2. Representations. (Value = 40%)<br />

Let B = {b 1 , b 2 , b 3 } be a basis of a vector space V . Let<br />

c 1 = b 1 + b 2 ,<br />

c 2 = b 2 + b 3 , <strong>and</strong><br />

c 3 = b 1 + b 3 .<br />

Let C = {c 1 , c 2 , c 3 }.<br />

(a) (Coordinates.) If [x] B =<br />

⎡<br />

⎣ −1 1<br />

−2<br />

⎤<br />

⎦, then what is x<br />

Solution: x = −b 1 + b 2 − 2b 3<br />

2

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