Linear Algebra, Theory And Applications, 2012a

494 ANSWERS TO SELECTED EXERCISES
25 (0, −1, 0) (4, −1, 0) saddle point. (2, −1, −12) local
minimum.
27 (1, 1) , (−1, 1) , (1, −1) , (−1, −1) saddle points.
( √ ) (
−
1
6√
5 6, 0 ,
1
√ )
6√
5 6, 0 Local minimums.
29 Critical points: (0, 1, 0) , Saddle point.
31 ±1
G.14 Exercises
8.4
1 The first three vectors form a basis and the dimension
is 3.
3 No. Not a subspace. Consider (0, 0, 1, 0) and multiply
by −1.
5 NO. Multiply something by −1.
7 No. Take something nonzero in M where say u 1 =
1. Now multiply by 100.
9 Suppose {x 1 , ··· , x k } is a set of vectors from F n .
Show that 0 is in span (x 1 , ··· , x k ) .
0 = ∑ i 0x i
11 ⎛It is a⎞
subspace. ⎛ ⎞ It ⎛is spanned ⎞ by
3 2 1
⎜ 1
⎟
⎝ 1 ⎠ , ⎜ 1
⎟
⎝ 1 ⎠ , ⎜ 0
⎟
⎝ 0 ⎠ . These are also independent
so they constitute a
0 0 1
basis.
13 Pick n points {x 1 , ··· ,x n } . Then let e i (x) = 0 unless
x = x i when it equals 1. Then {e i } n i=1
is linearly
independent, this for any n.
15 { 1,x,x 2 ,x 3 ,x 4}
17 L ( ∑ n
i=1 c iv i ) ≡ ∑ n
i=1 c iw i
19 No. There is a spanning set having 5 vectors and
this would need to be as long as the linearly independent
set.
23 No. It can’t. It does not contain 0.
43 Suppose ∑ n
i=1 a ig i =0. Then 0 = ∑ i a ∑
∑
i j A ijf j =
j f ∑
j i A ija i . It follows that ∑ i A ija i = 0 for
each j. Therefore, since A T is invertible, it follows
that each a i = 0. Hence the functions g i are linearly
independent.
G.15 Exercises
9.5
1 ThisisbecauseABC is one to one.
7 In the following examples, a linear transformation,
T is given by specifying its action on a basis β. Find
its matrix with respect to this basis.
( ) 2 0
(a)
1 1
( ) 2 1
(b)
1 0
( ) 1 1
(c)
2 −1
11 A =
13
⎛
⎜
⎝
⎛
⎜
⎝
0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0
1 0 2 0 0
0 1 0 6 0
0 0 1 0 12
0 0 0 1 0
0 0 0 0 1
⎞
⎟
⎠
⎞
⎟
⎠
15 You can see these are not similar by noticing that
the second has an eigenspace of dimension equal to
1 so it is not similar to any diagonal matrix which
is what the first one is.
19 This is because the general solution is y p + y where
Ay p = b and Ay = 0. Now A0 = 0 and so the solution
is unique precisely when this is the only solution
y to Ay = 0.
G.16 Exercises
25 No. This would lead to 0 = 1.The last one must
not be a pivot column and the ones to the left must
each be pivot columns.
10.6
( 1 1
2 Consider
0 1
Jordan form.
) ( 1 0
,
0 1
)
. These are both in