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Linear Algebra, Theory And Applications, 2012a

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G.10. EXERCISES 491<br />

11 It follows from the fact that e 1 , ··· , e m occur as<br />

columns in row reduced echelon form that the dimension<br />

of the column space of A is n and so, since<br />

this column space is A (R n ) , it follows that it equals<br />

F m .<br />

12 Since m>nthe dimension of the column space of<br />

A is no more than n and so the columns of A cannot<br />

span F m .<br />

15 If ∑ ∑ i c iz i = 0, apply A to both sides to obtain<br />

i c iw i = 0. By assumption, each c i =0.<br />

19 There are more columns than rows and at most m<br />

can be pivot columns so it follows at least one column<br />

is a linear combination of the others hence A<br />

is not one too one.<br />

21 |b−Ay| 2 = |b−Ax+Ax−Ay| 2<br />

= |b−Ax| 2 + |Ax − Ay| 2 +2(b−Ax,A (x − y))<br />

= |b−Ax| 2 +|Ax − Ay| 2 +2 ( A T b−A T Ax, (x − y) )<br />

= |b−Ax| 2 + |Ax − Ay| 2 and so, Ax is closest to b<br />

out of all vectors Ay.<br />

27 No.<br />

⎛<br />

⎜<br />

⎝<br />

1 0 2 0<br />

0 1 1 7<br />

0 0 0 1<br />

0 0 0 0<br />

⎞<br />

⎟<br />

⎠<br />

29 Let A be an m × n matrix. Then ker (A) is a subspace<br />

of F n . Is it true that every subspace of F n is<br />

the kernel or null space of some matrix? Prove or<br />

disprove.<br />

Let M be a subspace of F n . If it equals {0} , consider<br />

the matrix I. Otherwise, it has a basis {m 1 , ··· , m k } .<br />

Consider the matrix<br />

(<br />

m1 ··· m k 0 )<br />

where 0 is either not there in case k = n or has<br />

n − k columns.<br />

30 This is easy to see when you consider that P ij is<br />

its own inverse and that P ij multiplied on the right<br />

switches the i th and j th columns. Thus you switch<br />

the columns and then you switch the rows. This has<br />

the effect of switching A ii and A jj . For example,<br />

⎛<br />

⎜<br />

⎝<br />

1 0 0 0<br />

0 0 0 1<br />

0 0 1 0<br />

0 1 0 0<br />

⎞ ⎛<br />

⎟ ⎜<br />

⎠ ⎝<br />

a b c d<br />

e f z h<br />

j k l m<br />

n t h g<br />

⎞<br />

⎟<br />

⎠ ·<br />

⎛<br />

⎜<br />

⎝<br />

1 0 0 0<br />

0 0 0 1<br />

0 0 1 0<br />

0 1 0 0<br />

⎞ ⎛<br />

⎟<br />

⎠ = ⎜<br />

⎝<br />

a d c b<br />

n g h t<br />

j m l k<br />

e h z f<br />

⎞<br />

⎟<br />

⎠<br />

More formally, the ii th entry of P ij AP ij is<br />

∑<br />

P ij<br />

is A spP ij<br />

pi = P ij<br />

ij A jjP ij<br />

ji = A ij<br />

s,p<br />

31 If A has an inverse, then it is one to one. Hence the<br />

columns are independent. Therefore, they are each<br />

pivot columns. Therefore, the row reduced echelon<br />

form of A is I. This is what was needed for the<br />

procedure to work.<br />

G.10 Exercises<br />

5.8<br />

1<br />

⎛<br />

⎝ 1 2 2 1 0<br />

3<br />

⎞<br />

⎠ . =<br />

⎛<br />

1 2 3<br />

⎞ ⎛<br />

⎞<br />

1 0 0 1 2 0<br />

⎝ 2 1 0 ⎠ ⎝ 0 −3 3 ⎠<br />

1 0 1 0 0 3<br />

⎛ ⎞ ⎛ ⎞<br />

3 ⎝ 1 2 1<br />

1 2 2 ⎠ . = ⎝ 1 0 0<br />

0 0 1 ⎠ ·<br />

⎛<br />

2 1 1<br />

⎞ ⎛<br />

0 1 0<br />

⎞<br />

⎝ 1 0 0<br />

2 1 0 ⎠ ⎝ 1 2 1<br />

0 −3 −1 ⎠<br />

1 0 1 0 0 1<br />

⎛ ⎞ ⎛<br />

⎞<br />

1 2 1 1 0 0 0<br />

5 ⎜ 1 2 2<br />

⎟<br />

⎝ 2 4 1 ⎠ . = ⎜ 0 0 0 1<br />

⎟<br />

⎝ 0 0 1 0 ⎠ ·<br />

3 2 1 0 1 0 0<br />

⎛<br />

⎞ ⎛<br />

⎞<br />

1 0 0 0 1 2 1<br />

⎜ 3 1 0 0<br />

⎟ ⎜ 0 −4 −2<br />

⎟<br />

⎝ 2 0 1 0 ⎠ ⎝ 0 0 −1 ⎠<br />

1 0 −1 1 0 0 0<br />

⎛<br />

⎞<br />

9 ⎝ 1 2 1 0<br />

3 0 1 1 ⎠<br />

1 0 2 1<br />

⎛ √ √ ⎞<br />

1<br />

11 11<br />

1<br />

√ 11√<br />

√ 10<br />

√ 11 0<br />

√<br />

= ⎝ 3<br />

11 11 −<br />

3<br />

110 10 11 −<br />

1<br />

10√<br />

2 5 ⎠<br />

√ √ √ √ √<br />

·<br />

1<br />

11 11 −<br />

1<br />

110 10 11<br />

3<br />

10 2 5<br />

⎛ √ √ √<br />

11<br />

2<br />

11 11<br />

6<br />

11 11<br />

4<br />

√ √ √ √ 11√<br />

11<br />

√<br />

⎝<br />

2<br />

0<br />

11 10 11<br />

1<br />

22 10 11 −<br />

2<br />

√ √ 55√<br />

√ 10<br />

√ 11<br />

1<br />

0 0<br />

2 2 5<br />

1<br />

5 2 5<br />

⎞<br />

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