Homework 2 Solutions

Josh Hernandez
October 27, 2009
1.4 - Linear Combinations and Systems of Linear Equations
2. Solve the following systems of linear equations.
b.
2x
1
7x
2
+ 4x
3
= 10
x
1
2x
2
+ x
3
= 3
2x
1
x
2
2x
3
= 6
Solution:
1. Scaling down from rst pivot:
1 ( 2x
1
7x
2
+ 4x
3
= 10)
-2 ( x
1
2x
2
+ x
3
= 3)
-1 ( 2x
1
x
2
2x
3
= 6)
2. Summing down from rst row:
2x
1
7x
2
+ 4x
3
= 10
-3x
2
+ 2x
3
= 4
-6x
2
+ 6x
3
= 4
3. Scaling down from second pivot:
2x
1
7x
2
+ 4x
3
= 10
2( -3x
2
+ 2x
3
= 4)
-1( -6x
2
+ 6x
3
= 4)
4. Summing down from second row:
2x
1
7x
2
+ 4x
3
= 10
-3x
2
+ 2x
3
= 4
-2x
3
= 4
5. Scaling up from second pivot:
-3 ( 2x
1
7x
2
+ 4x
3
= 10)
7 ( -3x
2
+ 2x
3
= 4)
-2x
3
= 4
6. Summing up from second pivot:
-6x
1
+ 2x
3
= -2
-3x
2
+ 2x
3
= 4
-2x
3
= 4
7. Summing up from third pivot:
-6x
1
= 2
-3x
2
= 8
-2x
3
= 4
8. Solution:
x
1
= -1/3
x
2
= -8/3
x
3
= -2
d.
x
1
+ 2x
2
+ 2x
3
= 2
x
1
+ 8x
3
+ 5x
4
= -6
x
1
+ x
2
+ 5x
3
+ 5x
4
= 3
1
Solution:
1. Scaling down from rst pivot:
x
1
+ 2x
2
+ 2x
3
= 2
-1 ( x
1
+ 8x
3
+ 5x
4
= -6)
-1 ( x
1
+ x
2
+ 5x
3
+ 5x
4
= 3)
2. Summing down from rst row:
x
1
+ 2x
2
+ 2x
3
= 2
2x
2
+ -6x
3
5x
4
= 8
x
2
+ -3x
3
5x
4
= -1
3. Scaling down from second pivot:
x
1
+ 2x
2
+ 2x
3
= 2
2x
2
+ -6x
3
5x
4
= 8
-2 ( x
2
+ -3x
3
5x
4
= -1)
4. Summing down from second row:
x
1
+ 2x
2
+ 2x
3
= 2
2x
2
+ -6x
3
5x
4
= 8
5x
4
= 10
5. Scaling up from second pivot:
-1 ( x
1
+ 2x
2
+ 2x
3
= 2)
2x
2
+ -6x
3
5x
4
= 8
5x
4
= 10
6. Summing up from second pivot:
-x
1
+ -8x
3
5x
4
= 6
2x
2
+ -6x
3
5x
4
= 8
5x
4
= 10
7. Summing up from third pivot:
-x
1
+ -8x
3
= 16
2x
2
+ -6x
3
= 18
5x
4
= 10
8. Solution:
x
1
= -16 + -x
3
x
2
= 9 + 3x
3
x
3
= x
3
x
4
= 2
This solution set is the line through the point (-16, 9, 0, 2), in the direction (-1, 3, 1, 0).
3. For each of the following lists of vectors in R
3
, determine whether the rst vector can be expressed as
a linear combination of the other two.
b. (1, 2, -3), (-3, 2, 1), (2, -1, -1).
2
Solution: We can nd such a linear combination i.e. a, b R such that
a(1, 2, -3) + b(-3, 2, 1) = (2, -1, -1)
if we can nd a solution to the linear system
a + -3b = 2
2a + 2b = -1
-3a + b = -1
1. Scaling down from rst pivot:
6 ( a + -3b = 2)
-3 ( 2a + 2b = -1)
2 ( -3a + b = -1)
2. Summing down from rst row:
a + -3b = 2
-24b = 15
-16b = 10
The bottom two rows are equivalent.
3. Scaling up from the second pivot:
8 ( a + -3b = 2)
3 ( 8b = -5)
3. Summing up from the second row:
a = 1
8b = -5
4. Solution:
a = 1
b = -5/8
So these vectors are indeed linearly dependent.
d. (2, -1, 0), (1, 2, -3), (1, -3, 2).
Solution: Suppose there existed such a linear combination,
a(2, -1, 0) + b(1, 2, -3) = (1, -3, 2)
Then there must be a solution to the linear system
2a + b = 1
-a + 2b = -3
-3b = 2
So b = -2/3. Plugging this into the two rst equations,
2a + -2/3 = 1
-a + 2(-2/3) = -3
Solving for a gives conicting results: a = 5/6, and 5/3. This set of equations has no solution, so
the three vectors are linearly independent.
5h. Determine whether the given vector is in the span of S:
v =
_
1 0
0 1
_
, S =
_
v
1
=
_
1 0
-1 0
_
, v
2
=
_
0 1
0 1
_
, v
3
=
_
1 1
0 0
__
3
Solution: Suppose there existed a, b, c R such that av
1
+ bv
2
+ cv
2
= v. A quick check of the
set reveals that only v
1
has a nonzero element in the bottom-left position. Since v has a zero in
that position, then a = 0. Likewise, considering the bottom-right element, we see that b = 1.
This leaves us with the much simpler linear combination,
_
0 1
0 1
_
+ c
_
1 1
0 0
_
=
_
c 1 + c
0 1
_
=
_
1 0
0 1
_
.
Considering the elements, we get the conicting solutions c = 0 and c = -1.
10. Show that if
M
1
=
_
1 0
0 0
_
, M
2
=
_
0 0
0 1
_
, and M
2
=
_
0 1
1 0
_
then the span of {M
1
, M
2
, M
3
} is the set of all symmetric 2 2 matrices.
Solution: The generic symmetric 2 2 matrix satises the equation A = A
t
, or
_
a
11
a
12
a
21
a
22
_
=
_
a
11
a
21
a
12
a
22
_
This gives the solutions
a
11
= a
11
, a
12
= a
12
, a
21
= a
12
, a
22
= a
22
and the solution set
__
a
11
a
12
a
12
a
22
_
: a
11
, a
12
, a
22
R
_
.
This generic symmetric matrix can be generated from the given matrices,
a
11
_
1 0
0 0
_
+ a
12
_
0 1
1 0
_
+ a
22
_
0 1
1 0
_
,
so they do indeed span the symmetric matrices.
14. Show that if S
1
and S
2
are arbitrary subsets of a vector space V, then
span(S
1
S
2
) = span(S
1
) + span(S
2
).
Solution:
span(S
1
S
2
) =
_
n

i=1
c
i
u
i
:
i
S
1
S
2
, c
i
R
_
=
_
n

i=1
c
i
u
i
: u
i
S
1
or u
i
S
2
, c
i
R
_
.
Now, we can divide the u
i
into those vectors from S
1
and those from S
2
. Conversely, we can
combine the v
i
and w
i
into a sum of vectors from S
1
S
2
.
=
_
m

i=1
c
i
v
i
+
m

i=1
d
i
w
i
: v
i
S
1
, w
i
S
2
, c
i
, d
i
R
_
= {v + w : v span(S
1
), w span(S
2
)}
= span(S
1
) + span(S
2
).
4
1.5 - Linear Dependence and Linear Independence
2. Determine whether the following sets are linearly dependent or independent.
d. {x
3
x, 2x
2
+ 4, -2x
3
+ 3x
2
+ 2x + 6} in P
3
(R)
Solution: The linear equation
a(x
3
x) + b(2x
2
+ 4) + c(-2x
3
+ 3x
2
+ 2x + 6) = 0
can be rearranged, combining like terms,
x
3
(a 2c) + x
2
(b + 3c) + x(-a + 2c) + 1(4b + 6c) = 0.
Since dierent powers of x are linearly independent

, this linear combination must be trivial, i.e.

a 2c = 0
b + 3c = 0
-a + 2c = 0
4b + c = 0
The rst and third equations are obviously equivalent. Dropping the third, we have a system
of three equations, the rst of which is independent of the other two (it contains a variable not
found in the other two), which are independent of one another (they are not multiples of one
another). Such a system (three independent equations in three unknowns) has a unique solution.
Since (0,0,0) is obviously a solution, this system is linearly independent.
* (This comes from the denition of the ring of polynomials. However, it can also be derived
when one considers polynomials as real functions. If )
f. {(1, -1, 2), (2, 0, 1), (-1, 2, -1)} in R
3
Solution:
a + 2b + -c = 0
-a + 2c = 0
2a + b + -c = 0
Matlab gives only the trivial solution. These vectors are independent.
h.
__
1 0
-2 1
_
,
_
0 -1
1 1
__
-1 2
1 0
__
2 1
1 -2
__
in M
22
(R).
Solution: We can treat these matrices as 4-dimensional column vectors.
a + -c + 2d = 0
-b + 2c + d = 0
-2a + b + c + d = 0
a + b + -2d = 0
Matlab gives only the trivial solution. These vectors are independent.
4. In F
n
, let e
j
denote the vector whose jth coordinate is 1 and whose other coordinates are 0. Prove
that {e
1
, e
2
, . . . , e
n
} is linearly independent.
5
Solution: If a
j
F, then a
j
e
j
is the vector whose jth coordinate is a
j
and whose other coordinates
are 0. By our rules of vector addition, the linear combination a
1
e
1
+ a
2
e
2
+ + a
n
e
n
produces
the vector (a
1
, a
2
, . . . , a
n
) F
n
. If
a
1
e
1
+ a
2
e
2
+ + a
n
e
n
= 0 = (0, 0, . . . , 0),
then a
1
= a
2
= = a
n
= 0. The only valid linear combination is the trivial one, thus the vectors
are linearly independent.
9. Let u and v be distinct vectors in a vector space V. Show that {u, v} is linearly dependent if and only
if u or v is a multiple of the other.
Solution:
Suppose u and v are linearly dependent, that is, there exist a and b, at least one of which is
not zero, such that au+bv = 0. Assume a = 0 (otherwise swap u and v).

Then, we can rearrange

the equation, dividing both sides by a, to get
u = -
b
a
v.
Thus u is a multiple of v.
Suppose u = kv or v = ku for some k in R (if the latter case, swap u and v). Then
we can rearrange the equation,
u + (-k)v = 0
so u and v are linearly independent.
* This condition is important, since any algebraic proof requiring division by zero is no good. Swapping
u and v allows me to start fresh with a better arrangement.
15. Let S = {u
1
, u
2
, . . . , u
n
} be a nite set of vectors. Prove that S is linearly dependent if and only if
u
1
= 0 or u
k+1
span({u
1
, u
2
, . . . , u
k
}).
6
Solution:
Suppose S is linearly dependent. Then there exist a
1
, . . . , a
n
, at least one of which is nonzero,
such that
a
1
u
1
+ a
2
u
2
+ + a
n
u
n
= 0
Let k be the largest integer 1 k n such that a
k+1
= 0. If k = 0, we have the equation
a
1
u
1
= 0. Then u
1
= 0, since a
1
= 0. If otherwise, we have the equation
a
1
u
1
+ a
2
u
2
+ + a
k
u
k
+ a
k+1
u
k+1
= 0
(we can ignore the zero terms to the right of u
k+1
). Rearranging and dividing by a
k+1
= 0,
-a
1
a
k+1
u
1
+
-a
2
a
k+1
u
2
+ +
-a
k
a
k+1
u
k
= u
k+1
.
Thus u
k+1
lies in span({u
1
, . . . , u
k
}).
Suppose u
k+1
span({u
1
, u
2
, . . . , u
k
}). Then there exist a
1
, . . . , a
k
such that
u
k+1
= a
1
u
1
+ a
2
u
2
+ + a
k
u
k
.
Dene a
k+1
= -1, and a
i
= 0 for k + 2 i n. Then
a
1
u
1
+ a
2
u
2
+ + a
n
u
n
= a
1
u
1
+ + a
k
u
k
a
k+1
u
k+1
= 0,
so S is linearly independent.
18. Let S be a set of nonzero polynomials in P(F) such that no two have the same degree. Prove that S
is linearly independent.
Solution: Note that S may be innite. However, a linear combination is dened as a nite
weighted sum of vectors. We need only worry about n vectors at a time.
Consider a linear combination
a
1
f
1
+ a
2
f
2
+ + a
n
f
n
= 0
where f
i
are polynomials in S, and a
i
S. These f
i
must all have dierent degrees, so assume
they are ordered from smallest to largest degree (otherwise, we can rearrange them and start the
proof over). If the linear combination is nontrivial, let a
k+1
be the last nonzero coecient. If
f
k+1
has degree d, then a
k+1
f
k+1
can be written
a
k+1
c
0
+ a
k+1
c
1
x + + a
k+1
c
d
x
d
.
The polynomials f
1
, . . . , f
k
have degree strictly less than d, so a
i
f
i
has zero x
d
coecient for all
i = k.No combination of these polynomials could cancel out the x
d
term of f
k
. Therefore a
k
= 0,
so the linear combination must be trivial.
1.6 - Bases and Dimension
3. Determine which of the following sets are bases for P
2
(R).
b. {1 + 2x + x
2
, 3 + x
2
, x + x
2
}
7
Solution: Solving the linear system
a + b + c = 0
2a + c = 0
a + 3b = 0
,
Matlab gives only the trivial solution. Thus the vectors are linearly independent.
d. {-1 + 2x + 4x
2
, -2 + 3x x
2
, 1 x + 6x
2
}
Solution: Solving the linear system
4a + -b + 6c = 0
2a + 3b + -c = 0
-a + -2b + c = 0
,
Matlab gives only the trivial solution. Thus the vectors are linearly independent.
8. Let W denote the subspace of R
5
consisting of all the vectors having coordinates that sum to zero.
The vectors
u
1
= (2, -3, 4, -5, 2), u
2
= (-6, 9, -12, 15, -6)
u
3
= (3, -2, 7, -9, 1), u
4
= (2, -8, 2, -2, 6)
u
5
= (-1, 1, 2, 1, -3), u
6
= (0, -3, -18, 9, 12)
u
7
= (1, 0, -2, 3, -2), u
8
= (2, -1, 1, -9, 7)
generate W. Find a subset of the set {u
1
, u
2
, . . . , u
8
} that is a basis for W.
8
Solution: This problem is a great opportunity to exhibit a very economical linear-systems
solving technique.
First, we drop the second vector, which is obviously a multiple of the rst. The third
vector is not, so {u
1
, u
3
} is linearly independent. Well build up from that. The system
au
1
+ bu
3
= u
4
(notice the reordering) has 5 equations and 2 unknowns. To save time, Ill
rst solve the rst two equations. The (most likely unique) solution to that system can then be
tested in the rest of the equations.
2a + 3b = 2
-3a + -2b = -8
This has the unique solution a = 4, b = 2. Plugging these values in, we see that u
4
is indeed a
linear combination of u
1
and u
3
, so we drop it from the list. Next, try the same with u
5
:
2a + 3b = -1
-3a + -2b = 1
This gives the solution a = b = -
1
5
. Plugging these values in, we get
-
1
5
(2, -3, 4, -5, 2) + -
1
5
(3, -2, 7, -9, 1) = -
1
5
(5, -5, 11, -14, 3) = (-1, 1, 2, 1, -3) = u
5
Thus there is no solution to the linear combination - solving the rst two equations precludes
solving the rest. Thus {u
1
, u
3
, u
5
} is linearly independent. On to u
6
. Now we have a 3-by-3:
2a + 3b + -c = 0
-3a + -2b + c = -3
4a + 7b + 2c = -18
This gives the solution a = 1, b = -2, c = -4. Plugging these values in,
(2, -3, 4, -5, 2) + -2(3, -2, 7, -9, 1) + -4(-1, 1, 2, 1, -3) = (0, -3, -18, 9, 12) = u
6
So u
6
is a linear combination of {u
1
, u
3
, u
5
}. Moving on to u
7
,
2a + 3b + -c = 1
-3a + -2b + c = 0
4a + 7b + 2c = -2
This gives the solution a =
-13
21
, b =
8
21
, c =
-23
21
. Plugging that in, we get
-13
21
(2, -3, 4, -5, 2) +
8
21
(3, -2, 7, -9, 1) +
-23
21
(-1, 1, 2, 1, -3)
=
1
21
(21, 0, -42, -30, 21) = (1, 0, -2, 3, -2) = u
7
So {u
1
, u
3
, u
5
, u
7
} is linearly independent. We can stop here, if were smart. The space of vectors
in F
5
whose entries sum to 0 is a 4-dimensional space. Supposing weve picked the rst four
entries, the last one must be the negative of their sum - we have at most four degrees of freedom.
Another way to see this is that W is the null space of the linear transformation T : F
5
R, with
the mapping
T(v
1
, v
2
, v
3
, v
4
, v
5
) = v
1
+ v
2
+ v
3
+ v
4
+ v
5
The range of this linear transform is all of R (You can tweak the entries to sum to any real
number), so rank(T) = 1. Therefore
dim(W) = nullity(T) = dim(F
n
) rank(T) = 5 1 = 4.
9
We can design a nicer basis for this space by considering the generic vector,
_
a
1
, a
2
, a
3
, a
4
, -(a
1
+ a
2
+ a
3
+ a
4
)
_
.
Again, the rst four entries are free, but the last is the negative of their sum. Breaking this apart,
we get
a
1
(1, 0, 0, 0, -1) + a
2
(0, 1, 0, 0, -1) + a
3
(0, 0, 1, 0, -1) + a
4
(0, 0, 0, 1, -1),
so we have the basis
_
(1, 0, 0, 0, -1), (0, 1, 0, 0, -1), (0, 0, 1, 0, -1), (0, 0, 0, 1, -1)
_
10. In each part, use the Lagrange interpolation formula to construct the polynomial of smallest degree
whose graph contains the following points.
b. (-4, 24), (1, 9), (3, 3).
Solution: These points are actually co-linear. They lie on the line y = -3x + 12. The Lagrange
interpolator is
p(x) = 24
(x 1)(x 3)
(-4 1)(-4 3)
+ 9
(x -4)(x 3)
(1 -4)(1 3)
+ 3
(x -4)(x 1)
(3 -4)(3 1)
= x
2
_
24
35
+
9
-10
+
3
14
_
+ x
_
-4
24
35
+ 1
9
-10
+ 3
3
14
_
+
_
3
24
35
+ -12
9
-10
+ -4
3
14
_
= 0 3x + 12,
so just what we expected.
d. (-3, -30), (-2, 7), (0, 15), (1, 10).
Solution: This has Lagrange interpolator
p(x) = -30
(x+2)x(x1)
(-3+2)(-3)(-31)
+ 7
(x+3)x(x1)
(-2+3)(-2)(-21)
+ 15
(x+3)(x+2)(x1)
(3)(2)(-1)
+ 10
(x+3)(x+2)x
(1+3)(1+2)(1)
This simplies to (knowing that the function must pass through (0,15)),
x
3
(
-30
-12
+
7
6
+
15
-6
+
10
12
) + x
2
(1
-30
-12
+ 2
7
6
+ 4
15
-6
+ 5
10
12
) + x(-2
-30
-12
+ -3
7
6
+ 1
15
-6
+ 6
10
12
) + 15
= 2x
3
x
2
6x + 15
To test, plug in any of the x-values, and see that you get the right y-values.
12. Let u, v, w be distinct vectors of a vector space V. Show that if {u, v, w} is a basis for V, then
{u + v + w, v + w, w} is also a basis for V.
10
Solution: Suppose, for contradiction, that there existed some a, b, c, not all zero, such that
0 = a(u + v + w) + b(v + w) + cw = au + (a + b)v + (a + b + c)w.
Now, u, v, w are linearly independent, so all of these coecients have to be zero, i.e.
a = 0
a + b = 0
a + b + c = 0
This linear combination only has the trivial solution - in fact, swapping the a and c columns, and
then the rst and third rows, we have a system in row-echelon form:
c + b + a = 0
b + a = 0
a = 0
Thus a = b = c = 0, and so {u + v + w, v + w, w} are linearly independent.
15. The set of all n n matrices having trace equal to zero is a subspace W of M
nn
(F). Find a basis for
W. What is the dimension of W?
Solution: If A is such a matrix, one can freely choose every entry a
ij
up till a
nn
. This entry must
be equal to the negative sum of the diagonal entries a
11
, . . . , a
n1,n1
. A schematic of such a
matrix:
_
_
_
_
_
_
_
a
11
a
12
a
1,n-1
a
1n
a
21
a
22
a
2,n-1
a
2n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
n-1,1
a
n-1,2
a
n-1,n-1
a
n-1,n
a
n1
a
n2
a
n,n-1
-

n-1
i=1
a
ii
_
_
_
_
_
_
_
(1)
Breaking this apart into a basis, we have two dierent sorts of matrices; rst, those with zeros
along the diagonal, whose basis is
_

_
_
_
_
_
_
_
_
0 1 0 0
0 0 0 0
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
_
_
_
_
_
_
_
,
_
_
_
_
_
_
_
0 0 1 0
0 0 0 0
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
_
_
_
_
_
_
_
, . . . ,
_
_
_
_
_
_
_
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
0 0 0 0
0 1 0 0
_
_
_
_
_
_
_
,
_
_
_
_
_
_
_
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
0 0 0 0
0 0 1 0
_
_
_
_
_
_
_
_

_
There are n
2
n o-diagonal entries, so this basis contains n
2
n matrices. Second, there are
the diagonal matrices. These have the basis
_

_
_
_
_
_
_
_
_
1 0 0 0
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
0 0 0 -1
_
_
_
_
_
_
_
,
_
_
_
_
_
_
_
0 0 0 0
0 1 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 0
0 0 0 -1
_
_
_
_
_
_
_
, . . . ,
_
_
_
_
_
_
_
0 0 0 0
0 0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 1 0
0 0 0 -1
_
_
_
_
_
_
_
_

_
.
11
There are n1 of these basis matrices, one for each diagonal entry except the last. Joining these
two bases we have a linearly independent set (for each (i, j) = (n, n), there is exactly one basis
matrix M such that M
ij
= 0) spanning W (any matrix of the form in (1) can be built from a
combination of these vectors, where the coecients a
ij
correspond to the basis vectors in the
obvious way). The dimension of W is exactly (n
2
n) + (n 1) = n
2
1.
20. Let V be a vector space having dimension n, and let S be a generating subset of V.
(a) Prove that there is a subset of S that is a basis of V.
Solution: Let = {v
1
, . . . , v
n
} be a basis of V. We can not assume that S is nite, but we need
only nd a nite subset of S that generates .
For each v
i
, there is some collection {w
i1
, w
i2
, . . . , w
ik
i
} S and scalars d
i1
, . . . , d
ik
i
such that v
i
=

k
i
j=1
d
ij
w
ij
. Dene
T = {w
ij
: 1 i n, 1 j k
i
}
Then T is a nite subset of S. Observe that, for any v V , we can write v =

c
i
v
i
, so
v =
n

i=1
c
i
_
_
k
i

j=1
d
ij
w
ij
_
_
Thus T spans V . By Theorem 1.10, there must be a subset of T S which is a basis of V.
(b) Prove that S contains at least n vectors.
Solution: By the above, T S contains a basis of V, which must have exactly n vectors. Therefore
S must have at least n vectors.
33. Let W
1
and W
2
be subspaces of a vector space V with bases
1
and
2
, respectively.
(a) Assume W
1
W
2
= V. Prove that
1

2
= and
1

2
is a basis for V.
Solution: First, if v lies in
1

2
, then v W
1
W
2
= {0}. However, 0 can not be a member of
any basis, neither
1
or
2
. Therefore
1

2
= .
Denote the elements of
1
and
2
by v
1
, v
2
. . . and w
1
, w
2
, . . ., respectively. We now must
prove that (1)
1

2
is linearly independent, and that (2)
1

2
spans V.
(1) Suppose
c
1
v
1
+ + c
m
v
m
+ c
m+1
w
1
+ . . . + c
m+n
w
n
= 0 (2)
for some c
1
, . . . , c
m+n
. We can rearrange this sum
c
1
v
1
+ + c
m
v
m
= -c
m+1
w
1
+ . . . + -c
m+n
w
n
Now, the LHS lies in W
1
, and the RHS lies in W
2
, so both must lie in W
1
W
2
= {0} Thus
c
1
v
1
+ + c
m
v
m
= 0 and -c
m+1
w
1
+ . . . + -c
m+n
w
n
= 0
but both of these linear combinations must be trivial, since
1
and
2
are both bases. Thus
c
1
= c
2
= = c
m+n
= 0, so (2) must be trivial. Therefore
1

2
must be linearly independent.
12
(2) Now, given u V, we can write u = v + w, with v W
1
and w W
2
. Likewise, we can write
v =

n
1
i=1
c
i
v
i
and w =

n
2
i=1
c
i
w
i
. So
u = v + w =
n
1

i=1
c
i
v
i
+
n
2

i=1
c
i
w
i
,
which is a linear combination in
1

2
. So the union of bases spans V.
(b) Conversely, if
1

2
= and
1

2
is a basis for V, prove that W
1
W
2
= V.
Solution: Denote the elements of
1
and
2
by v
1
, v
2
. . . and w
1
, w
2
, . . ., respectively. We need
to prove that (1) W
1
W
2
= {0}, and (2) W
1
+W
2
= V.
(1) Suppose u W
1
W
2
. Then we can write u in two ways:
c
1
v
1
+ c
2
v
2
+ + c
m
v
m
= v = d
1
w
1
+ d
2
w
2
+ + d
n
w
n
.
This gives the linear combination
c
1
v
1
+ c
2
v
2
+ + c
m
+ -d
1
w
1
+ -d
2
w
2
+ + -d
n
w
n
= 0
Observe that all vectors in the sum lie in
1

2
, which is a basis, so all coecients must be
zero. Therefore u = 0. So W
1
W
2
= {0}.
(2) Now, any u V can be written as a linear combination in
1

2
, that is to say, of
vectors from either
1
or
2
. Rearrange this sum so that it has the form
u =
n
1

i=1
v
i
+
n
2

j=1
w
i
= v + w,
where v is some vector in W
1
and w is some vector in W
2
. Thus V = W
1
+W
2
.
13