Solution to Set 6, Math 2568
S 3.3: No. 5. For S = {a, d}, as given below, show that either Sp S = R2
or give an algebraic specification for Sp S. If Sp S 6= R2 , also give a geometric
specification of Sp S.
1
1
a=
, d=
−1
0
Solution: Take y = [y1 , y2 ]T ∈ R2 and inorder to describe the span of S set
1 1 x1
y1
x1 a + x2 d = y implying
=
−1 0 x2
y2
Augmented matrix of the above linear system on row reduction as shown below
gives
1 1 y1
1 1
y1
1 0
−y2
R2 +R1
R1 −R2
⇒
⇒
−1 0 y2
0 1 y2 + y1
0 1 y2 + y1
It is clear that there is no restriction for y = [y1 , y2 ]T so that the linear system
can be solved for [x1 , x2 ]T . Hence Sp S = R2 in this case.
S 3.3: No. 18. For S = {v, w, z}, as given below, show that either SpS = R3
or give an algebraic specification for Sp S. If Sp S 6= R3 , also give a geometric
specification of Sp S.
 
 
 
1
0
1
v = 2 , w = −1 , z = 0
0
1
2
Solution: Take y = [y1 , y2 , y3 ]T ∈ R3 . We seek x = [x1 , x2 , x3 ]T so that
x1 v + x2 w + x3 z = y. This implies
   

y1
x1
1 0 1
2 −1 0x2  = y2 
y3
x3
0 1 2
Augmented matrix of the above linear system on row reduction as shown below
gives

1
2
0
0
−1
1


1 y1
1
0 y2  ⇒R2 −2R1 0
2 y3
0
0
−1
1



y1
1 0 1
y1
y2 − 2y1  ⇒−R2 0 1 2 −y2 + 2y1 
y3
0 1 2
y3


1 0 1
y1
−y2 + 2y1 
⇒R3 −R2 0 1 2
0 0 0 y3 + y2 − 2y1
1
−2
2
Since the above corresponds to a linear system for x that only has solution if
y is restricted with y3 + y2 − 2y1 = 0, it is clear that we need the Sp S =
1
y ∈ R3 : y3 + y2 − 2y1 = 0 , which is the equation of a plane normal to the
vector (−2, 1, 1)T . Alternately, if we take y1 = t, y2 = s, arbitrary, we have
T
T
T
y3 = 2t − s and so y = [t,
s, 2t − s] = t[1, 0, 2] + s[0, 1, −1] and so Sp S =
T
T
Sp [1, 0, 2] , [0, 1, −1] = Sp {v, w}.
S 3.3: No. 36. Same question as No. 32, but with


1 0 −1
A = −1 1 2 
1 2 2
Solution: As in problem 32 above, we set Ax = y, now with both x =
[x1 , x2 , x3 ]T , y = [y1 , y2 , y3 ]T ∈ R3 , in order that matrix multiplication makes
sense. Then, row reduction of the linear system leads to






1 0 −1 y1
1 0 −1
y1
1 0 −1
y1
−1 1 2 y2  ⇒R2 +R1 0 1 1 y2 + y1  ⇒R3 −2R2 0 1 1

y2 + y1
R3 −R1
0 0 1 y3 − 3y1 − 2y2
1 2 2 y3
0 2 3 y3 − y1




1 0 −1
y1
1 0 0 −2y1 + y3 − 2y2
⇒R2 −R3 0 1 0 3y2 + 4y1 − y3  ⇒R1 +R3 0 1 0 3y2 + 4y1 − y3 
0 0 1 y3 − 3y1 − 2y2
0 0 1 y3 − 3y1 − 2y2
Clearly, the system of equations is consistent regardless of y and so R(A) = R3 .
T
Also, if we set y = 0 above
the solution set for x is just3x = 0 = [0, 0, 0] and
3
hence N (A) = 0 ∈ R . (Note we are denoting 0 ∈ R since the zero vector
has a different number of component zeros for different Rn ).
S 3.3: No. 38 For A as given below, determine for b = [1, −3]T , if i. b ∈ R(A).
ii. If b ∈ R(A), exhibit a vector x ∈ R2 such that Ax = b. iii. If b ∈ R(A),
then write b as a linear combination of columns of A.
1 −2
A=
−3 6
Solution: We seek to solve Ax = [1, −3]T which gives rise to the augmented
matrix below, which on row reduction gives
1 −2 1
1 −2 1
R2 +3R1
⇒
−3 6 −3
0 0 0
which is consistent and has solution x1 = 1 + 2t, x2 = t, arbitrary. Therefore,
b = [1, −3]T ∈ R(A). To answer ii, we note that x = [x1 , x2 ]T = [1 + 2t, t]T =
[1, 0]T +t[2, 1]T . So, in particular with choice t = 0, we have x = [1, 0]T satisfying
Ax = b as also follows from inspection. To answer part iii. With x = [1, 0]T
choice, we have clearly x1 A1 + x2 A2 = Ax = [1, −3]T and we therefore have
b = A1 , which is clear from inspection of the first column of matrix A and the
vector b.
S 3.3, No. 40. Same question as (38.) except now A is as given below


1 −2 1
A = 2 −3 5
1 0 7
2
and different cases of b as listed below
 
 
 
 
 
 
1
1
4
0
0
0
(i)b = 2 , (ii)b =  1  , (iii)b = 7 , (iv)b = 1 , (v)b =  1  , (vi)b = 0
0
−1
2
2
−2
0
Solution: To answer the question for a whole lot of different b, it is better
to keep b = [b1 , b2 , b3 ]T and determine conditions on b and then check if those
conditions hold for each of the b provided. Now, if we set Ax = b, it is clear
that x = [x1 , x2 , x3 ]T ∈ R3 and the augmented matrix for the resulting linear
system row reduces as shown:

1 −2 1
[A|b] = 2 −3 5
1 0 7

1 −2 1
⇒R3 −2R2 0 1 3
0 0 0



b1
1 −2 1
b1
2 −2R1 
0 1 3 b2 − 2b1 
b2  ⇒ R
R3 −R1
0 2 6 b3 − b1
b3



b1
1 0 7
−3b1 + 2b2
b2 − 2b1  ⇒R1 +2R2 0 1 3
b2 − 2b1 
b3 + 3b1 − 2b2
0 0 0 b3 + 3b1 − 2b2
It is clear that b ∈ R(A) if and only if b3 +3b1 −2b2 = 0. With given b, we readily
check this is satisfied for cases (ii), (iii), (iv), (vi), in which case b ∈ R(A), but
the condition is not satisfied in cases (i), (v). x in case (ii) may be chosen (with
x3 = 0) to be [−3b1 +2b2 , b2 −2b1 , 0]T = [−3+2, 1−2, 0]T = [−1, −1, 0]T and we
can readily verify on inspection that indeed −A1 − A2 = b. Now consider case
(iii) which for x3 = 0 leads to x = [−3(4) + 2(7), 7 − 2(4), 0]T = [2, −1]T and we
may readily verify that 2A1 − A2 = b in this case. Now consider case (iv). We
have for x3 = 0, x = [−3b1 + 2b2 , b2 − 2b1 , 0]T = [−3(0) + 2, 1 − 2(0)]T = [2, 1]T
and accordingly we can readily verify that 2A1 + A2 = b. The last case, case
(vi) is trivial since b = [0, 0, 0]T = 0A1 + 0A2 + 0A3 .
S 3.3: No. 42 For W as below exhibit a 3 × 3 matrix A so that W = R(A)
and conclude that W is a subspace.
W = x = [x1 , x2 , x3 ]T : 3x1 − 4x2 + 2x3 = 0
We note that the description of W , implies x2 = s and x3 = t are arbitrary
and thus, x1 = 34 s − 23 t. Therefore, for x ∈ W , we have x = [x1 , x2 , x3 ]T =
T
T
s 34 , 1, 0 + t − 23 , 0, 1 and so
(
T T )
4
2
W = Sp
, 1, 0 , − , 0, 1
3
3
. and therefore we may choose
4
− 32
0
1
3
A = 1
0
3

0
0
0
Note R(A) is the column space of A, which is the span of the column vectors.
Since the zero vector in the third column is not independent of the first two
columns, the span of columns of A is W and therefore equal to R(A). Since
R(A) is a subspace, it follows that W is a subspace of R3 .
S 3.3: No. 44 Let S = {v, w, x} as given below is a set of vectors. Exhibit a
matrix A such that Sp S = R(A).
 
 
 
1
0
1
v = 2 , w = −1 , , x =  1 
0
1
−1
Solution: From Matrix multiplication we know that for a 3 × 3 matrix A and
x = [x1 , x2 , x3 ]T , Ax = x1 A1 + x2 A2 + x3 A3 ; therefore, span of the column
vectors of A is the same as the range of A. Therefore, choosing A1 = v, A2 = w
and A3 = x will ensure that R(A) = Sp S, and so


1 0
1
A = 2 −1 1 
0 1 −1
S 3.4 No. 2. Let W be the subspace of R4 so consisting of vectors x =
[x1 , x2 , x3 , x4 ]T which satisfy the following relations
x1 + x2 − x3 + x4 = 0
x2 − 2x3 − x4 = 0
Find a basis for W .
Solution: Clearly, this corresponds to the augmented matrix system which
row-reduces in the form shown
1 1 −1 1 0
1 0 1
2 0
⇒R1 −R2
0 1 −2 −1 0
0 1 −2 −1 0
implying that we may choose x3 = s, x4 = t as arbitrary and then x1 =
T
T
−s − 2t, x2 = 2s + t and therefore x = [−s − 2t, 2s + t, s, t] = s [−1, 2, 1, 0] +
T
T
T
t [−2, 1, 0, 1] . This means that W = Sp [−1, 2, 1, 0] , [−2, 1, 0, 1] . Further, it is clear that this set of two vectors are independent since if we set
c1 [−1, 2, 1, 0]T + c2 [−2, 1, 0, 1]T = [0, 0, 0, 0]T it follows by just looking at last
two components of thevectors on the left and right
hand side of that c1 = 0
and c2 = 0. Therefore [−1, 2, 1, 0]T , [−2, 1, 0, 1]T is a basis for W .
S 3.4, No. 10. For x as given below, determine if x ∈ W , where W is the
subspace in Problem 2 above. If x ∈ W , express x as a linear combination of
the basis vectors found in Exercise 2 above.
(a) x = [−3, 3, 1, 1]T , (b.) x = [0, 3, 2, −1]T , (c.) x = [7, 8, 3, 2]T , (d.) x = [4, −2, 0, −2]T
Solution: To answer all the questions at the same time, it is better to express
x = [x1 , x2 , x3 , x4 ]T and then set x = c1 u1 + c2 u2 with vectors {u1 , u2 } =
4
[−1, 2, 1, 0]T , [−2, 1, 0, 1]T , which form a basis vector for subspace W as shown
in exercise 2. We will then find conditions on [x1 , x2 , x3 , x4 ]T that ensures that
x ∈ W and then check if these are satisfied for the specific x given in cases (a)(d) above. The linear system for unknown (c1 , c2 ) clearly has the augmented
matrix which reduces as shown:






1 2
−x1
−1 −2 x1
1 2 −x1


2


1 x2 

 ⇒−R1 2 1 x2  ⇒R2 −2R1 0 −3 x2 + 2x1 
R
−R

1



3
1
0 −2 x3 + x1 
0 x3
1 0 x3
0 1
x4
0
1 x4
0 1 x4




1 2
−x1
1 2
−x1


0 1
x4
x4 

 R3 +2R2 0 1
⇒R2 →R4 ,R4 →R2 
0 −2 x3 + x1  ⇒R4 +3R2 0 0 x3 + x1 + 2x4 
0 0 x2 + 2x1 + 3x4
0 −3 x2 + 2x1


1 0
−x1 − 2x4
0 1

x4

⇒R1 −2R2 
0 0 x3 + x1 + 2x4 
0 0 x2 + 2x1 + 3x4
It follows that for x ∈ W we need x3 + x1 + 2x4 = 0 and x2 + 2x1 + 3x4 = 0,
and in the case when these two conditions are satisfied, we have c1 = −x1 −
2x4 and c2 = x4 . Now, in case a, x3 + x1 + 2x4 = 1 − 3 + 2(1) = 0 and
x2 + 2x1 + 3x4 = 3 + 2(−3) + 3(1) = 0 hence [−3, 3, 1, 1]T ∈ W . In case b,
x3 + x1 + 2x4 = 2 + 0 + 2(−1) = 0 and x2 + 2x1 + 3x4 = 3 + 2(0) + 3(−1) = 0 and
so [0, 3, 2, −1]T ∈ W . In case c. x3 + x1 + 2x4 6= 0, and hence [7, 8, 3, 2]T ∈
/ W.
In case d., we have x3 +x1 +3x4 = 0+4+3(−2) 6= 0. Hence [4, −2, 0, −2]T ∈
/W
either.
S 3.4: No. 12. For matrix A given below, Find a. a basis B in reduced
row echelon form such that B is equivalent to A. b. Find a basis for the null
space of A. c. Find a basis for the range of A that consists of columns of A.
For each column Aj that does not appear in the basis, express Aj as a linear
combination of the basis vectors. d. Find a basis for the row space of A.


1 1 2
A = 1 1 2
2 3 5
Solution: First check the null space, i.e. Find the solution set
augmented linear system for this on row reduction gives:





1 1 2 0
1
1 1 2 0
1 1 2 0 ⇒R2 −R1 0 0 0 0 ⇒R2 →R3 ,R3 →R2 0
R3 −2R1
2 3 5 0
0 1 1 0
0
of Ax = 0. The


0
1
0 ⇒R1 −R2 0
0
0
(1)
Therefore, we have x3 = t arbitrary, x1 = −t and x2 = −t.
Therefore
x = t[−1, −1, 1]T and a null space basis is simply [−1, −1, 1]T , and we have
5
1
1
0
2
1
0
0
1
0
1
1
0

0
0
0
answered b.. Also, from the row reduction above, ignoring the zeros of the
augmented matrix [A|0] above, it is clear that A row-reduces to B which is in
the row reduced echelon form and is given by


1 0 1
B = 0 1 1
0 0 0
and we have answered part a. To answer part c we note from null space computation that x3 = t, x1 = −t = x2 , If we picked t = 1, we would have the
statment Ax = 0 read as x1 A1 + x2 A2 + x3 A3 = 0 = −A1 − A2 + A3 . Therefore A3 = A1 + A3 and we can take {A1 , A2 } = {[1, 1, 2]T , [1, 1, 3]T } as basis
for R(A) since it has the same span as {A1 , A2 , A3 } and from row reduction
process in (1) without the third column, it is clear that x1 = 0, x2 = 0 and the
set is linearly independent. So, we have answered part c. Now, for part d, we
note that row reduction does not change the row space. So, the nonzero rows
of B will be a basis for the row space of A, i.e. {[1, 0, 1], [0, 1, 1]} is a basis for
row space of A.
S 3.4: No. 26 Find a basis for the range of each of the matrices below
1 0 0
1 1 0
1 1 0
a. A =
, b. A =
, c. A =
1 0 1
1 1 0
1 1 1
We note easily that in cases a, b one of the columns of A is zero. Since any
set with 0 vector cannot be linearly indepedendent, we discard this
column, We
also note that the remaining nonzero columns are [1, 1]T , [0, 1]T in case a are
linearly independent since they consist of nonparallel vectors. We can also show
the same by setting c1 [1, 1]T + c2 [0, 1]T = [0, 0]T and showing (c1 , c2 ) = (0, 0)
as the only possibility.
In
the case b. since two nonzero columns are identical,
it is clear that [1, 1]T is a basis for R(A). In
case c since, two columns
are indentical, and it is clear that [1, 1]T , [0, 1]T are nonparallel vectors and
therefore linearly independent, we have this as a basis for R(A) since the column
space of A is the same as R(A) and column space is not affected by throwing
out vectors that may be written in terms of other vectors.
S 3.4: No. 30 If B = {v1 , v2 , v3 } is a set of three indepedent vectors in R3
show that B is a basis for R3 .
Solution: We have to show that the Sp B = R3 , i.e. for any y ∈ R3 we can
find x1 , x2 , x3 so that x1 v1 + x2 v2 + x3 v3 = y. We denote the augmened system
of this linear system as [A|y]. Since three vectors are involved in R3 , it follows
that A is a 3 × 3 matrix. From linear independence of the columns of A (which
are vj ), we know that if y = 0 then the only solution is (x1 , x2 , x3 ) = 0. This
implies A row reduces to I and therefore, on row reduction [A|y] will reduce to
[I|A−1 y] and we will obtain solution x for any y ∈ R3 without restriction on y.
Hence B is a basis for R3 .
S 3.4, No. 37. Prove that every basis in R2 contains exactly two vectors.
Proceed by showing the following:
a. A basis for R2 cannot have more than two vectors.
6
b. A basis for R2 cannot have one vector.
Solution: a. Suppose have have more than two vectors in a basis for R2 , say
n vectors {v1 , v2 , · , vn } with n > 2 in a basis. Now check linear dependence
or independence by setting x1 v1 + x2 v2 + · · · + xn vn = 0. The augmented
matrix for the corresponding linear system is [A|0] where A is a matrix whose
column vectors are v1 , v2 , · · · , vn . Since each vector is in R2 while there are
n such columns, it is clear that A is a non-square matrix with of size 2 × n
which on row reducation to rref form will have columns without a leading one,
since the number of leading ones cannot exceed the number of rows which is 2,
where as there are n columns involved. This means that x1 , x2 , x2 , · · · , xn need
not be all zero, implying that the set {v1 , v2 , · · · , vn } cannot be independent
contradicting the fact that this set is supposed to be a basis. Therefore the
initial assumption n > 2 must be incorrect and n ≤ 2.
b. Now, we want to rule out n < 2. Assume n = 1. In that case we have
only one basis vector {v1 }. Let’s say v1 = [a, b]T . Since v1 is a basis element,
v1 6= 0, implying either a or b is nonzero. If a 6= 0, b = 0, then clearly [0, 1]T
cannot be expressed as a multiple of v1 . Similarly if b 6= 0, a = 0, then [1, 0]T ∈
/
Sp {v1 }. If both a, b 6= 0 then clearly [2a, b]T is not parallel to [a, b]T and hence
[2a, b]T ∈
/ Sp {v1 }. In every case there is a vector which is not expressible in
terms of v1 no matter what v1 looks like. Therefore, number n of vectors in
any basis for R2 cannot be 1. From part a. we conclude that n = 2
S 3.5, No. 8: Determine by inspection if S = {v1 , v3 } forms a basis for R3 ,
where v1 = [1, −1, 1]T and
−1, 0]T .
v3 = [1,
T
Solution We know that [1, 0, 0] , [0, 1, 0]T , [0, 0, 1]T span R3 (any vector being a linear combination of this three) and also independent set since x1 [1, 0, 0]T +
x2 [0, 1, 0]T + x3 [0, 0, 1]T = [0, 0, 0]T implies x1 , x2 , x3 = 0. Therefore, dimR3 =
3. The set S has two vectors and therefore cannot be a basis.
S 3.5, No. 12: Determine if S = {v1 , v2 , v3 } is a basis for R3 by using Theorem
9 property 3, where v1 = [1, −1, 1]T , v2 = [0, 1, 2]T , and v3 = [1, −1, 0]T .
Solution: According to the theorem, we only need to show show that S is
linearly independent. For that purpose we first set x1 v1 + x3 v2 + x3 v3 = 0 and
ask the question if [x1 , x2 , x3 ] has to be 0 or not. But this amounts to looking
for solution of a linear system of equation whose augmented matrix row-reduces
as shown below






1 0 1 0
1 0 1 0
1 0 1 0
−1 1 −1 0 ⇒R2 +R1 0 1 0 0 ⇒R3 −2R2 0 1 0 0
R3 −R1
1 2 0 0
0 2 −1 0
0 0 −1 0




1 0 1 0
1 0 0 0
⇒−R3 0 1 0 0 ⇒R1 −R3 0 1 0 0
0 0 1 0
0 0 1 0
which implies x1 = 0 = x2 = x3 and therefore the set of vectors S is linearly
independent. and hence from Thm 9, part 3, S is a basis for R3 .
S 3.5, No. 20: Determine dimW , where W is the set of vectors x =
7
[x1 , x2 , x3 , x4 ]T that satisfy the relation below:
x1 − x2 + 0x3 + 0x4 = 0
0x1 + x2 − 2x3 + 0x4 = 0
0x1 + 0x2 + x3 − x4 = 0
Solution: Since x ∈ W ⊂ R4 satisfies the above linear system of equation, we
look at the corresponding augmented matrix and row reduce so find how the
solution set x may be described. Row reduction gives as shown:





1 −1 0
0 0
1 0 −2 0 0
1 0 0 −2
0 1 −2 0 0 ⇒R1 +R2 0 1 −2 0 0 ⇒R2 +2R3 0 1 0 −2
R1 +2R3
0 0
1 −1 0
0 0 1 −1 0
0 0 1 −1

0
0
0
The above RREF matrix contains the information for the linear system where
x1 − 2x4 = 0, x2 − 2x4 = 0, x3 − x4 = 0, with x4 = t arbitrary. Therefore,
x1 = 2x4 = 2t, x2 = 2x4 = 2t, x3 = x4 = t, implying that for any
x ∈ W , we
have x = [x1 , x2 , x3 ]T = t[2, 2, 1, 1]T . Therefore, it is clear that [2, 2, 1, 1]T
forms a basis for W and therefore dim W = 1.
S 3.5, No. 22: Find a basis for N (A) for the given matrix A and give the
nullity and rank of A:
−1 2 0
A=
2 −5 1
Solution: First to determine N (A), we have to find the solution set of vectors
x ∈ R3 satisfying Ax = 0. The augmented matrix for this linear system rowreduces as shown:
−1 2 0 0
1 −2 0 0
1 −2 0 0
1
−R1
R2 −2R1
−R2
[A|0] =
⇒
⇒
⇒
2 −5 1 0
2 −5 1 0
0 −1 1 0
0
This immediately implies x1 − 2x3 = 0, x2 − x3 = 0. We can set x3 = t, and
we then have x1 = 2x3 = 2t and x2 = x3 = t. Therefore for
any x ∈ N (A),
we have the representation x = t[2, 1, 1]T and hence S = [2, 1, 1]T forms
a basis for N (A) for A as given above. Therefore, nullity of A is 1 (recall
nullity is the number of vectors in a basis for N (A)). From theorem about
Nullity (A) + rank (A) = n = 3, number of columns of A, we have rank A =
3 − 1 = 2. Alternately, if we did not want to use this theorem, we can argue
from the above row reduction that if we were to solve Ax = y for any y ∈ R2 ,
since the row reduced form of A has no zero rows, there will be no
restriction on
y ∈ R2 and hence R(A) = R2 which has dimension = 2 since [1, 0]T , [0, 1]T
is a basis for R2 .
S 3.5, No. 24: Same question as No. 22, except for the matrix


1 2 0 5
A = 1 3 1 7
2 3 −1 9
8
−2
1
0
−1
0
⇒R1 +2
0
Solution: As before, to determine N (A), we have to find the solution set of
vectors x ∈ R4 (note A in this case has four columns, unlike previous problem)
satisfying Ax = 0. The augmented matrix for this linear system row-reduces as
shown:




1 2 0 5 0
1 2
0
5 0
2 −R1
0 1
1
2 0
[A|0] = 1 3 1 7 0 ⇒R
R3 −2R1
2 3 −1 9 0
0 −1 −1 −1 0




1 0 −2 1 0
1 0 −2 0 0
R2 −2R3 
1 −2R2 
0 1 1 2 0 ⇒R
0 1 1 0 0
⇒R
R3 +R2
1 −R3
0 0 0 1 0
0 0 0 1 0
This implies that x1 − 2x3 = 0, x2 + x3 = 0, x4 = 0 with x3 = t arbitrary.
Therefore, x1 = 2x3 = 2t, x2 = −x3 = −t and x = t[2, −1, 1, 0]T . This is
true for arbitrary x ∈ N (A). Therefore, [2, −1, 1, 0]T spans N (A) and since
only
one nonzero
vector is involved trivially linearly independent. Therefore
[2, −1, 1, 0]T forms a basis for N (A) and therefore nullity of A is again 1.
Since number of columns of A is four, it follows from theorem that rank A =
4 − nullity A = 4 − 1 = 3. (Just a side comment: Note that this implies that
R(A) = R3 since there are three vectors in a basis for R(A) and any linearly
independent vectors in R3 forms a basis for R3 as well). We could have got to
the same conclusion by considering Ax = y and from noticing that the rref [A]
has no nonzero rows it would have followed that rref [A|y] has no nonzero rows
in the coefficient part and therefore no restriction on y ∈ R3 .
9