Page 329 - 35Linear Algebra

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(b) This is a vector space. Although, the question does not ask you to, it is
a useful exercise to verify that all ten vector space rules are satisfied.
(c) This is not a vector space for many reasons. An easy one is that
(1, −1, 0) and (−1, 1, 0) are both in the space, but their sum (0, 0, 0) is
not (i.e., additive closure fails). The easiest way to repair this would
be to drop the requirement that there be at least one entry equaling 1.

9. (i) Thanks to multiplicative closure, if u ∈ U, so is (−1)·u. But (−1)·u+u =
(−1)·u+1·u = (−1+1)·u = 0.u = 0 (at each step in this chain of equalities
we have used the fact that V is a vector space and therefore can use its vector
space rules). In particular, this means that the zero vector of V is in U and
is its zero vector also. (ii) Also, in V , for each u there is an element −u
such that u+(−u) = 0. But by additive close, (−u) must also be in U, thus
every u ∈ U has an additive inverse.

 
x
y
(a) This is a vector space. First we check additive closure: let   and
0
       
z x z x + z
w be arbitrary vectors in U. But since y + w = y + w ,
       
0 0 0 0
so is their sum (because vectors in U are those whose third component
 
x
y
vanishes). Multiplicative closure is similar: for any α ∈ R, α   =
0
 
αx

 αy , which also has no third component, so is in U.
0
(b) This is not a vector space for various reasons. A simple one is that
   
1 2
0
u =   is in U but the vector u + u =  0  is not in U (it has a 2
z 2z
in the first component, but vectors in U always have a 1 there).

10.
     
1 1 −1 2 1 0 0 0 1 1 −1 2
1 3 2 2 1 1 0 0 0 2 3 0
     
  =    
−1 −3 −4 6 −1 0 1 0 0 −2 −5 8
     
0 4 7 −2 0 0 0 1 0 4 7 −2
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