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```					         MATH10212 Linear Algebra • Examples 5 • SOLUTIONS

As usual, the sequence of e.r.o.s is not unique.

1. It is a subspace: [0, 0] ∈ S; [0, y1 ] + [0, y2 ] =                     rank is 3, not 2, as it would be if it was possible.
[0, y1 + y2 ] ∈ S, so S is closed under +; and                             So w ∈ row(A).
c[0, y] = [c · 0, cy] = [0, cy] ∈ S, so S is closed
under multiplication by scalars. (Alternatively,              5. We take the transposes: then decide if the
this is the null space of the homogeneous system              system with augm. matrix [AT |wT ] is consis-
                              
1 1 −1                 1 1 −1
x = 0.)                                                                               R3 +R1            R −2R2
tent:  0 1 1  −→ 0 1 1  3              −→
[3.5.3] It is a subspace: [0, 0] = [0, 2 · 0] ∈ S;                    −1 1 1                  0 2 0

[x1 , 2x1 ] + [x2 , 2x2 ] = [(x1 + x2 ), 2(x1 + x2 )] ∈ S, so   1 1 −1
S is closed under +; and c[x, 2x] = [cx, 2cx] ∈ S, 0 1 1  r.e.f., we have a bad row, so incon-
so S is closed under multiplication by scalars.                 0 0 −2
(Alternatively, this is the null space of the ho- sistent, so wT ∈ col(AT ), so w ∈ row(A).
mogeneous system 2x − y = 0.)                                                                       
−1
2. [3.5.5] It is a subspace: [0, 0, 0] ∈ S;                                             1 0 −1           0
6. [3.5.15] That is, is             · 3 =      ?
[x1 , x1 , x1 ] + [x2 , x2 , x2 ]                                                       1 1 1              0
−1
= [x1 + x2 , x1 + x2 , x1 + x2 ] ∈ S, so S is closed                                           0
under +; and c[x, x, x] = [cx, cx, cx] ∈ S, so S is Left-hand side is actually = 1 , so the answer
closed under multiplication by scalars. (Alterna- is “no”.
tively, this is the null space of the homogeneous
system x − y = 0; x − z = 0.)                                 7. (a) “Three-in-one”: we user.e.f. for all:
                                             
1 0 −1 1                          1 0 −1 1
R2 −R1      R3 −R1
[3.5.7] It is not a subspace: e.g. [0, 0, 0] ∈ S.             1 1 1 1 −→ ... −→ 0 1 2 0
1 2 3 1                         0 2 4 0
3. Geometrically, clearly, the sum of two vec-
tors on the given line is again on that line, R3 −2R2                   1 0 −1 1
as is a scalar multiple, by deﬁnitions. Or:                     −→ 0 1 2 0 = R, r.e.f. The non-zero
S = {c[a1 , a2 , a3 ] | c ∈ R}; then [0, 0, 0] =                        0 0 0 0
0[a , a , a ] ∈ S; c [a , a , a ] + c [a , a , a ] =          rows form a basis of row(A):
1    2   3             1   1   2    3        2   1   2    3
(c1 + c2 )[a1 , a2 , a3 ]       ∈   S;       d(c[a1 , a2 , a3 ])   =       {[1 0 − 1 1],    [0 1 2 0]}.
(dc)[a1 , a2 , a3 ] ∈ S.
Leading ones are in 1st and 2nd columns of
4.       b ∈ col(A) ⇔ the lin.     system is consis-                       R; hence these columns form a basis of col(R);
1 0 −1 3                                           columns of A form a basis
hence the 1st and 2nd 
tent with the augmented matrix
1 1 1 2                                         1      0 
R2 −R1 1 0 −1        3                                                     of col(A):   1 , 1 .
−→                     , r.e.f., no bad row, so con-                                           
0 1 2 −1                                                                         1     2
sistent, so b ∈ col(A).                            We use r.e.f. of A to ﬁnd the solution of Ax = 0,
We add third row w and try to       make it zero which is x2 = −2x3 , x1  x − x4 with x3 , x4

= 3       ,
1 0 −1                              x1      s−t
by e.r.o.s “from top downwards”:  1 1 1  being free variables: x2  =  −2s  We give to
           
−1 1    1                          x3   s 
                                                         x4        t
1 0 −1             1 0 −1
−→ ... −→ 1 0 1 2  −→ 2 0 1 2 ; it the free var. the values 1, 0 and 0, 1, and thus
R2 −R1   R3 +R               R3 −R

0 1 0              0 0 −2
is kind of clear that this cannot be done: in fact

1
MATH10212 • Linear Algebra • Solutions of Examples 5                                                    2

                                                         
 1
                   −1      1 −1 0                    1 −1 0
                 0    −1 0              R2 +R1
0 −1 1  R−→ 2  3 +R
−2 ,             .                 1     −→
obtain a basis of null(A):
 1 

 0 
   0      1  −1              0 1 −1
                       
0                 1       1 −1 0
0 −1 1; r.e.f. The non-zero rows form a
(b) “Three-in-one”: we use r.e.f.
                                       for all:

1 1 0        1           1 1 0        1               0 0 0
0 1 −1 1  R−→ 2 0 1 −1 1  = R;
3 −R                              basis of the span; reverting to columns: a basis:
   
0 1 −1 −1                0 0 0 −2                     1          0 
−1 , −1 .
r.e.f. The non-zero rows form a basis of row(A):                      
0        1
{[1 1 0 1], [0 1 − 1 1], [0 0 0 − 2]}.
(b) We form the matrix of these rows:
Leading entries are in 1st, 2nd, and 4th columns                                  
2 −3 1             1 −1 0
of R; hence these columns form a basis of col(R);                     2 ↔R
1 −1 0 R−→ 1 2 −3 1 R2 −2R1 ... R3 −4R1
−→    −→
hence the 1st, 2nd, and 4th columns of A form a
     
 4 −4 1            4 −4 1
 1      1       1 
basis of col(A):   0 , 1 ,  1  .                   1 −1 0
                                    0 −1 1; r.e.f. The non-zero rows form a ba-
0     1      −1
0 0 1
We use r.e.f. of A to ﬁnd (by back subst.) the         sis of the span:
solution of Ax = 0, which is x4 = 0,
1   −1   0 ,   0   −1   1 ,   0 0   1   .
−x
x2 = x3 − x4 = x3 , x1 = −x2 − x4 = 3 with x3
      
x1      −t                    (In fact, here dimension is 3, so this is the whole
x     t                     space R3 ; any other basis would do, say, the
being a free variable:  2  =   We give to
x3    t                     standard one.)
x4       0                    10.    We work with the matrix formed by
             
the 
the free var. value 1 and thus obtain a basis                                    1 −1 0
 −1 
                                         these vectors as columns: −1 0
R2 +R
1  −→ 1
   
1 .                                                               0
of null(A):
  1                                                                 1
 −1

      
                                    1 −1 0               1 −1 0
0                                                      3 +R
0 −1 1  R−→ 2 0 −1 1 = R; r.e.f.
                                 
1 1 1                       1 1 1            0 1 −1               0 0 0
 0 1 2 R3 +R1      R4 −R1 0 1 2          Leading entries are in 1st and 2nd columns of
8. AT =            
−1 1 3 −→ ... −→ 0 2 4
      
R; hence these columns form a basis of col(R);
hence the 1st and 2nd columns of A form a basis
 1 1 1                        0 0 0
of   (which is the span of the given vectors):
1 1 1                                          col(A)   
R3 −2R2 0 1 2                                         1        −1     
−→            
0 0 0; r.e.f. The non-zero rows form            −1 ,  0  , .
                 
0 0 0                                             0       1
a basis of row(AT ); taking    
 their transposes we        11. (a) rank = 2; nullity = 2.
 1       0 
obtain a basis of col(A): 1 , 1 . (It is a         (b) rank = 3; nullity = 1.
            
1      2                  12. One explanation is by Th. 2.8: any set of
mere coincidence that the answer is the same as        m vectors in Rn is linearly dependent if m > n;
obtained by “2nd Method” in Q7(a); a basis is          here we have ﬁve vectors in R3 . Another expla-
never unique; and these two methods in general         nation is by the Rank Th. 3.24: row rank is equal
produce different bases.)                              to column rank; here we cannot have row rank
9. (a) It is easier to work with rows, so we take      greater than 3, so the columns cannot be lin. in-
transposes and form the matrix of the resulting        dep.
rows:                                                  13. Rank can be anything from 0 (for the zero
matrix) to 3, so nullity, = 5 – rank, can be any-
thing from 5 to 2.
MATH10212 • Linear Algebra • Solutions of Examples 5                                                 3

14.
                                                     16. We need to solve the linear system
1    2 a
−2 4a 2 R2 +2R1 ... R3 −aR1
−→          −→                       x1 b1 + x2
          b2 = w.  Augmented matrix:

a −2 1                                                1 1 1                  1 1 1
2 0 6        R2 −2R1
−→     0 −2 4 R3 −(1/2)R2
−→
                       
1      2         a                                     0 −1 2                0 −1 2
0 4a + 4 2a + 2 R3 +(1/2)R2                          
−→                         1 1 1
0 −2a − 2 1 − a2                                     0 −2 4; r.e.f., no bad row, so consis-
                                                       0 0 0
1     2          a
0 4a + 4                                              tent, so w ∈ span(B). Solution is x2 = −2,
2a + 2  =
3
0     0      2 + a − a2                              x1 = 1 − x2 = 3. Thus, [w]B =    .
−2
                             
1      2              a                          17. (a) By “Helpful Hint”, the rows of AB are
= 0 4(a + 1)          2(a + 1) .                    linear combinations of the rows of B. Then
0      0       (a + 1)(2 − a)                    any linear combination of the rows of AB, af-
                        ter substitution of those linear combinations of
1 2 −1
If a = −1, then this is = 0 0 0 , rank = 1.          the rows of B and collecting terms, is again
0 0 0                    a linear combination of the rows of B. Thus,
                          row(AB) ⊆ row(B). By Theorem 3.23+, then
1 2 2                     dim row(AB) ≤ dim row(B), which is the same
If a = 2, then this is = 0 12 6, rank = 2.           as rank(AB) ≤ rank(B).
0 0 0
(b) the simplest is with A = O.
In all other cases the diagonal elements are non-
zero, so rank is 3.                                    18.    By Q17 applied twice, rank(A) =
                                          rank(U −1 U A) ≤ rank(U A) ≤ rank(A), so we
1 1 0                1 1 0                   must have equality throughout.
15.
R2 −R1
1 0 1 −→ 0 −1 1 R−→ 2           3 +R

0 1 1                0 1 1                   19. The ith column of A2 = A · A is Aai ,
                                                     where ai is the ith column of the right factor
1 1 0
0 −1 1; r.e.f.; rank = 3 (we know that row           A. Since A2 = O, we have Aai = 0, so all
0 0 2                                                columns of A are in null(A). Since null(A) is
rank = column rank), so they do form a basis of        a subspace and therefore closed under taking
R3 by Th. 3.27.                                        linear combinations, all the linear combinations
                                                 of the ai are also contained in null(A), that
1 1 1 0                                           is, col(A) ⊆ null(A). By Theorem 3.23+ then
1 1 0 1 R2 −R1 R3 −R1                            dim col(A) ≤ dim null(A), that is, rank(A) ≤
(b)             
1 0 1 1 −→ ... −→                                nullity(A). Together with the Rank–Nullity
0 1 1 1                                           equality rank(A) + nullity(A) = n, this implies
                                                   rank(A) ≤ n .
2
1 1      1 0              1 1     1 0
0 0 −1 1 R2 ↔R3 0 −1 0 1 R4 +R2
                                      
0 −1 0 1 −→ 0 0 −1 1 −→
0 1      1 1              0 1     1 1
                                      
1 1      1 0              1 1     1 0
0 −1 0 1 R4 +R3 0 −1 0 1
                                      
0 0 −1 1 −→ 0 0 −1 1; r.e.f.,
0 0      1 2              0 0     0 3
rank = 4; hence they do form a basis of R4 by
Th. 3.27

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