Notes on Homework 6
1.
(a) We need to demonstrate three properties:
nonempty Since U is a subspace, 0 ∈ U; since W is a subspace, 0 ∈ W. Therefore,
0 ∈ U ∩ W, and U ∩ W is nonempty.
closed under addition Suppose v1 , v2 ∈ U ∩ W. Then in particular v1 , v2 ∈ U; since U
is a subspace, v1 + v2 ∈ U. Similarly, since W is a subspace, v1 + v2 ∈ W. Therefore,
v1 + v2 ∈ U ∩ W.
closed under multiplication Suppose v ∈ U ∩ W, λ ∈ F. Since v ∈ U and U is a
subsapce, λv ∈ U. Similarly, λv ∈ W. Therefore, λv ∈ U ∩ W.
x
2
(b) Many examples are possible. For instance, let V = R ; U = {
: x ∈ R }; W =
0
0
x
{
: y ∈ R}. Then U ∪ W consists of all vectors
such that (at least) one of x
y
y
and y is zero. Since
1
0
,
∈ U∪W
0
1
but
1
0
+
0
1
=
1
1
6∈ U ∪ W,
U ∪ W is not closed under addition, and thus isn’t a subspace.
2. The example given in 1b is closed under scalar multiplication, although it isn’t a vector
space. (Indeed, let U and W be any two subspaces of a vector space V. Suppose v ∈ U ∪ W
and λ ∈ F; we want to show λv ∈ U ∪ W. By definition of the union, either v ∈ U or v ∈ W.
If v ∈ U, then λv ∈ U, since U is a subspace; and U ⊆ U ∪ V, so that λv ∈ U ∪ W. Similarly,
if v ∈ W, then λv ∈ W ⊆ U ∪ W.)
3. Answers here are a little cursory.
(a) This is a subspace. Again, three things to check, presented here without much annotation.
i. Nonempty: (0, 0, 0) ∈ S.
ii. Addition: (0, a1 , b1 ) + (0, a2 , b2 ) = (0, a1 + a2 , b1 + b2 ) ∈ S.
iii. Multiplication: λ(0, a, b) = (0, λ a, λb) ∈ S.
(b) Not a subspace. Here are three ways of showing this. Any one is enough to show that
S isn’t a subspace of V!
i. (0, 0, 0) is not of the form (1, a, b), so the zero vector isn’t contained in S.
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
ii. (1, 2, 3), (1, 3, 5) are in S, but (1, 2, 3) + (1, 3, 5) = (2, 5, 8) 6∈ S.
iii. 2(1, 2, 3) = (2, 4, 6) 6∈ S.
(c) Not a subspace. For instance, the zero function z : x 7→ 0, which is the identity element
of V, is not in S. (Can also make examples by adding two functions together, e.g.,
f ( x) = g( x) = 3, but ( f + g)( x) = 6 so that ( f + g)(4) = 6 6= 3.)
(d) This is a subspace.
i. The zero function z satisfies z(4) = 0.
ii. If f , g ∈ S, then ( f + g)(4) = f (4) + g(4) by definition, but f (4) = g(4) = 0, so
( f + g)(4) = 0, and f + g ∈ S.
iii. If λ ∈ R, f ∈ S, then (λ f )(4) = λ · f (4) = λ · 0 = 0; λ f ∈ S.
(e) It’s a subspace.
i. The zero function z satisfies this differential equation.
ii. Suppose f , g ∈ S. Then
( f + g)′′ − ( f + g)′ = f ′′ + g′′ − f ′ − g′
= ( f ′′ − f ′ ) + ( g′′ − g′ )
= 0 − 0 = 0.
iii. Suppose λ ∈ R, f ∈ S. Then
(λ f )′′ − (λ f ′ ) = λ f ′′ − λ f ′
= λ( f ′′ − f ′ )
= λ · 0 = 0.
(f) It’s a subspace.
i. | z| ∈ S.
ii. If f , g ∈ S, α ∈ [−1, 1], then
( f + g)(−α ) = f (−α ) + g(−α )
= f (α ) + g(α )
= ( f + g)(α )
so f + g ∈ S.
iii. If f ∈ S, λ ∈ R, then for any α ∈ [−1, 1], (λ f )(−α ) = λ f (−α ) = λ f (α ) = (λ f )(α );
λ f ∈ S.
4.
(a) This doesn’t span; for instance, (1, 1, 2) T is not a multiple of (1, 1, 1) T .
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
(b) Spans. For any vector ( x, y, z), we want to find numbers a1 , a2 and a3 such that a1 (1, 2, 0) T +
a2 (0, 1, 0) T + a3 (0, 2, 1) T = ( x, y, z) T . This is tantamount to solving the system of equations
1 · a1 + 0 · a2 + 0 · a3 = x
2 · a1 + 1 · a2 + 2 · a3 = y
0 · a1 + 0 · a2 + 1 · a3 = z
We solve for a1 , a2 and a3 , representing the system with an augmented matrix:


1 0 0 x
 2 1 2 y 
0 0 1 z


1 0 0
x
 0 1 2 y − 2x 
0 0 1
z


x
1 0 0
 0 1 0 y − 2x − 2z 
z
0 0 1
Therefore, a1 = x, a2 = y − 2x − 2z, a3 = z should work. (Check this!)
(c) Spans. The vectors given here contain the vectors from the previous part, and any set
which contains a spanning set spans. Make sure you can prove this!
(d) Try the same thing as in part (b), get an augmented matrix:


1
2 2 3 x
 2
3 2 4 y 
−2 −2 0 −2 z


1 2
2
3
x
 0 −1 −2 −2 y − 2x 
0 2
4
4 z + 2x


x
1 2
2
3

 0 −1 −2 −2
y − 2x
0 0
0
0 z − 2x + 2y
This means that ( x, y, z) T is in the span of this set of vectors if and only if z − 2x + 2y =
0. In particular, (0, 0, 1) T is not in the span.
(e) A subset of a set which fails to span can’t span. Make sure you can prove this! Again,
(0, 0, 1) T is not in the span of this set.
5. In general, to prove that two sets A and B are equal, show that A ⊆ B – that is, every element
of A is an element of B – and B ⊆ A - every element of B is an element of A.
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
First, we show that span(v1 , v2 ) ⊆ span(λv1 , v2 ). Suppose w ∈ span (v1 , v2 ); by definition,
there are numbers a1 , a2 ∈ F such that
w = a1 v1 + a2 v2 .
Since λ ∈ F is nonzero, we can divide by it; and in particular, a1 =
a1
λ λ v1 .
Then
w = a1 v1 + a2 v2
a
= 1 · λ v1 + a2
λ
∈ span(λv1 , v2 ),
since we’ve written w as a linear combination of λv1 and v2 .
Second, suppose u ∈ span(λv1 , v2 ); we’ll show that u ∈ span(v1 , v2 ). By definition, there
are numbers b1 , b2 ∈ F such that
u = b1 · λ v1 + b2 v2
= ( b1 λ ) v1 + b2 v2
∈ span(v1 , v2 ).
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008