Download Homework Solution Section 2.3 8. Applying Theorem 2.4, we check

Homework Solution
Section 2.3
8. Applying Theorem 2.4, we check to see if the system Ax = b is consistent. This is because, the theorem
says that this system is consistent iff b is a linear combination of the columns of the matrix A. This
is exactly what is required to show that b is in the span of the columns of A. By row reduction, we
could reduce the augmented matrix [A|b] to


28
1 0 −1 | − 3 


0 1 2 | 29 

3 
0 0 0 |
0
and we see that the system is consistent (actually has infinite solutions.) So, the vector b is in the
span of the columns of the matrix A.
!
3
0
Extra Show that, <2 = span
,
.
−2
1
3
Soln. We must show that x [ −2
] + y [ 01 ] = [ ab ] can always be solved. By row reduction, we can see that
x = a/3, y = (2a + 3b)/3, so for any choice of a, b, we have
2a + 3b 0
a 3
[ −2 ] +
[ 1 ] = [ ab ]
3
3
(check this!)
0
3
14. Describe the span of the vectors
,
, (a) geometrically and (b) algebraically.
0
4
Soln.(a) The set of all linear combinations of [ 00 ] and [ 34 ] is just the line through the origin with [ 34 ] as
the direction vector. Note that the vector [ 00 ] is simply 0 [ 34 ].
(b) The vector equation of a line with direction vector [ 34 ] is [ xy ] = t [ 34 ] . This means that the vector [ xy ]
x
y
is in the span of [ 34 ]. Solving this vector equation, we get x = 3t, y = 4t. So, = → 4x − 3y = 0.
3
4
So, the span of the given vectors is the line with equation 4x − 3y = 0.
 
 
1
−1
16(a) Geometrically, we can see that the set of all linear combinations of  0  and  1  is just the
−1
0
plane through the origin with these vectors as direction vectors. Since the third vector is just the
difference between the second and the first vectors, it does not affect the span.
 
 
 
x
1
−1
(b) Algebraically, the vector equation of the plane is y  = s  0  + t  1 . This means the vector
z
−1
0
 
x
y  is in the span of these two vectors. To get the general equation of this plane, use the system
z
s − t = x
t = y The augmented matrix reduces to give the
arising out of the vector equations,
−s
= z
equation x + y + z = 0.
18. u, v, w are in span(u, v, w), since u, v, w can be written as a linear combination of u, v, w as u =
1u + 0v + 0w, etc.
 
 
 
3
2
1





Extra. Determine if the vectors x = 2 , u = 1 , v = 2  are linearly independent. If they are
2
3
−6
dependent, find a dependence relation.
Soln. We want to find scalars c1 , c2 , c3 such that
 
 
   
2
3
1
0
c1 2 + c2 1 + c3 −5 = 0
1
2
2
0
Form the corresponding linear system and the augmented matrix and row reduce to get


1 0 −4 | 0
0 1 3 | 0
0 0 0 | 0
This gives the solution, c1 = 4c3 , c2 = −3c3 , the vectors are linearly dependent as we can find a
non-trivial solution for our linear combination. A possible depedence relation is v = −4x + 3u.
 
 
 
3
2
1





24. We wnat to see if the system α 2 + β 1 + γ 2  = 0 has a trivial or nonrivial solution. Let
2
3


 −6
3
2
1
1 1/2 1



1
2 has the REF 0 1 −4 after the elementary operations R1 ↔ R2 ; R2 ←
A = 2
2 3−6
0 0
0
R2 − 3R1 ; R3 ← R3 − R1 ; R3 ← R3 − 4R2 . So, the homogeneous system has infinitely many solutions
given by α = −3γ, β = 4γ, γ ∈ <. Hence the three vectors
are linearly
One particular
 
  dependent.
 
3
2
1
dependence relation is obtained by taking γ = 1 to give 3 2 − 4 1 =  2 
2
3
−6
26. The given vectors are dependent. This can be determined by inspection because the system Ax = 0
has nontrivial solution. This can be seen by noting that this is a homogeneous system with three
equations and 4 variables so the rank of the matrix A can be at most 3 and hence the number of free
variables must be at least 1. See Theorem 2.6 in section 2.2.
Another way is to see that we have four vectors in <3 showing that the vectors must be dependent.
13
To find a specific relationship, solving the system x1 u1 +x2 u2 +x3 u3 +x4 u4 = 0 gives x1 = x4 , x2 =
37
6
45
− x4 , x3 = − x4 is the solution. Setting x4 = 37, we get x1 = 13, x2 = −6, x3 = −45, x4 = 37 as
37
37
one non-trivial set of coefficients that will result in a linear combination of zero vector.
42(a) We will use the theorems of Sections 2.2 and 2.3. We are given that
P the columns of the matrix
A, call these a1 · · · an , are independent. This is true iff the system
ci vi = 0 has only the trivial
solution. So, the number of free variables in the solution to the system must be 0. So, the rank
of the matrix A must be n since the rank theorem says that number of free variables = number of
variables - rank(A).
(b) Let vi be the ith row of A. From Theorem 2.7, the vectors are linearly independent iff rank of A is
greater than or equal to n. But a has n rows so rank(A)≤ n. So, we have n ≤ rank(A) ≤ n ⇒
rank(A) = n.
47. Suppose that S = {v1 , · · · vk , v} is a set of vectors in <n and that v is a linear combination of
v1 · · · vk . If S 0 = {v1 , · · · , vk }, prove that Span(S) = Span(S’).
Proof: Since Span(S) contains all linear combinations of the vectors in S and since S’ is a subset of S,
every linear combination of vectors in S’ is also a linear combination of vectors in S. So, Span(S’) ⊂
Span(S). If u ∈ Span(S), then u is a linear combination of vectors in S. If this linear combination
does not involve the vector v, then u is a linear combination of only the vectors v1 , · · · , vk and hence
is a member of Span(S’). If the linear combination involves the vector v, since v is itself a linear
combination of v1 , · · · , vk , we could rewrite the linear combination without v. In that case also, u
is a linear combination of vectors in S’ and so is a member of Span(S’). So, in either case, Span(S)
⊂ Span(S’). Hence Span(S) = Span(S’).