Download Math 365 Mid-Term Exam Sollutions 1. Prove or give a counter

Math 365
Mid-Term Exam Sollutions
1. Prove or give a counter-example: in any vector space V , if (u1 , u2 ) is linearly independent and u3 is not a multiple of u1 and u3 is not a multiple of u2 , then (u1 , u2 , u3 ) is
also linearly independent.
We give a counter-example. Let V := R2 , and let u1 := (1, 0), u2 := (0, 1), and let
u3 := (1, 1). Notice that any multiple of u1 must have second term equal to 0, which
means that neither u2 or u3 are multiples of u1 . Similarly, any multiple of u2 must
have first term equal to 0, and so neither u1 or u3 are multiples of u2 . In particular:
(u1 , u2 ) is linearly independent, and u3 is not a multiple of either u1 or u2 . However,
(u1 , u2 , u3 ) is linearly independent, since u1 + u2 − u3 = (0, 0).
2. Prove or give a counter-example: if u1 , u2 , u3 are vectors in a vector space V and
span (u1 , u2 , u3 ) = span (u1 , u2 ) ⊕ span (u3 ), then (u1 , u2 , u3 ) is linearly independent.
We give a counter-example: let V := R2 , and let u1 = (1, 0), u2 := (0, 0), and let u3 :=
(0, 1). Since u2 is the zero vector, the (u1 , u2 , u3 ) is linearly dependent. It remains only
to show that span (u1 , u2 , u3 ) = span (u1 , u2 ) ⊕ span (u3 ). We will use Proposition 1.45,
which means that we need only show that span (u1 , u2 , u3 ) = span (u1 , u2 ) + span (u3 )
and span (u1 , u2 ) ∩ span (u3 ) = {~0}.
Showing span (u1 , u2 , u3 ) = span (u1 , u2 )+span (u3 ): Suppose first that x ∈ span (u1 , u2 , u3 ).
By definition, that means that x = a1 u1 + a2 u2 + a3 u3 for some a1 , a2 , a3 ∈ R.
Letting u = a1 u1 + a2 u2 ∈ span (u1 , u2 ) and w = a3 u3 ∈ span (u3 ), we see that
x = u + w ∈ span (u1 , u2 ) + span (u3 ). Therefore, span (u1 , u2 , u3 ) ⊆ span (u1 , u2 ) +
span (u3 ). Suppose next that x ∈ span (u1 , u2 ) + span (u3 ). That means x = u + w
for some u ∈ span (u1 , u2 ) and some w ∈ span (u3 ). Notice: u = a1 u1 + a2 u2 and
w = a3 u3 for some a1 , a2 , a3 ∈ R, and so x = a1 u1 + a2 u2 + a3 u3 ∈ span (u1 , u2 , u3 ).
Thus, span (u1 , u2 ) + span (u3 ) ⊆ span (u1 , u2 , u3 ). Thus, we have span (u1 , u2 , u3 ) =
span (u1 , u2 )+span (u3 ). (Notice: this part has nothing to do with our particular choice
of vectors u1 , u2 , u3 !)
Showing span (u1 , u2 ) ∩ span (u3 ) = {~0}: Suppose that x ∈ span (u1 , u2 ) ∩ span (u3 ).
That means that x = a1 u1 + a2 u2 and x = a3 u3 for some a1 , a2 , a3 ∈ R. In particular,
that means that x = a1 (1, 0) + a2 (0, 0) and x = a3 (0, 1). Therefore, we must have
(a1 , 0) = (0, a3 ), which means that a1 = 0 and a3 = 0. Since a3 = 0, we see that
x = a3 u3 = (0, 0).
3. Give an example of two non-trivial subspaces U and W of R3 such that R3 = U ⊕ W.
You must prove that your example does what you say it does! (Note: non-trivial means
that you subspaces cannot be either {~0} or all of R3 .) You may not use the vectors
(1, 0, 0), (0, 1, 0), (0, 0, 1).
Notice: by homework #5, problem #2, any list of three linearly independent vectors
will work!
4. Prove or give a counter-example: if (u1 , u2 , . . . , um ) and (w1 , w2 , . . . , wm ) are linearly
independent lists of vectors in a vector space V , then (u1 + w1 , u2 + w2 , . . . , um + wm )
is also linearly independent.
Counter-example: pick any linearly independent list (u1 , u2 , . . . , um ), and let wi = −ui .
Then u1 + w1 = ~0 and so (u1 + w1 , u2 + w2 , . . . , um + wm ) has a zero term and so must
be linearly independent.
A more concrete example: let V = R2 , and let u1 = (1, 0) and u2 = (0, 1). Then
(u1 , u2 ) is linearly independent, since (u1 , u2 ) is the standard basis of R2 . Next, let
w1 = (−1, 0) and let w2 = (0, −1). Notice that any multiple of w1 must have 0 in the
second position, and so w2 is not a multiple of w1 . Similarly, any multiple of w2 must
have 0 in the first position, and so w1 is not a multiple of w2 . Thus, by a homework
assignment, we know that (w1 , w2 ) is also linearly independent. Finally, notice that
(u1 + w1 , u2 + w2 ) = (~0, ~0), and the list (~0, ~0) is linearly dependent.