Experience is another word for mistakes.

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MATH 221 - Finding a unique row space solution. October 23, 2015

Experience is another word for mistakes.

1 2 3 x1

2

The general solution to the system Axg =

245

x2

=

1

is obtained from

x3

the row reduction 1 2 3 2 1 2 3 2 . This row reduction shows that

2451

0 0 -1 -3

1 0

a basis for the row space is RA = span

2 ,

0

. Further, from the row

3

-1

reduction x3 = 3 and x1 = -2x2 - 3x3 + 2 = -2x2 - 7. Hence, the general solution xg, in vector form, reads

-2 -7

xg

=

x2

1

+

0

x2

+

xp

(1)

0

3

where x2 is arbitrary. The form of the general solution shows N (A) = span {}. The Orthogonal Subspace Decomposition Theorem (OSDT (pronounced ohst) is on page 146) guarantees that there is a solution xr = + r where N (A) and r RA. In addition, these vectors are orthogonal, ? r = 0, and they vectors are unique. Equating xg and xr and solving for the unknown row space solution yields

r = (x2 - ) + xp RA.

(2)

Using the orthogonality of RA with N (A) leads to the equation 0 = ? r = ? {(x2 - ) + xp} = (x2 - )||||2 + ? xp = (x2 - )5 + 14

which implies x2 - = -14/5. Substituting this into (2) shows

-14 -2 -7

r = (x2 - ) + xp =

5

1 + 0

0

3

28/5 - 7 -7/5

0

= -14/5 = -14/5 + (36/5) 0

3

-21/5

1

1

0

=

(-7/5) 1 + (36/5) 0

RA.

3

1

If you wish to make "doubly sure" you have a particular solution, check that 2

Ar = 1 . To find xr or xg (they are the same through the connection

2

x2 = - 14/5), write

-2

-2 -7

xg = x2 + xp

=

(

-

14/5)

+

xp

=

1

- 14/5

1

+

0

0

0

3

-2

1

0

= 1 +

(-7/5) 1 + (36/5) 0

= + r = xr.

0

3

1

*****************************************************

Repeat the above for Axg =

123 246

x1

x2

=

x3

1 2 3 1 1 2 3 2 . This gives

2462

0000

1 . The row reduction reads 2

-2

-3 1

xg =

x2

1

+

x3

0

+

0

x2

+ x3

+ xp

0

1

0

where x2 and x3 are arbitrary. The form of the general solution implies

1

N (A) = span ,

and the row reduction shows that RA = span

2

. Once

3

again, OSDT guarantees a solution xr = 1 + 2 + r where 1 + 2 N (A) and r RA. These vectors are orthogonal so that they satisfy 1 + 2 ? r = 0. Equating these two solutions and solving for the unknown row space solution gives

r = (x2 - 1) + (x3 - 2) + xp

(3)

Using the orthogonality (RA N (A)), it follows that 0 = r ? = r ? . This leads to two equations in the unknows x2 - 1 and x3 - 2. Solve to find x2 - 1 = 1/7 and x3 - 2 = 3/14 which, when substituted into (3) gives

-2

-3 1 -2/7 - 9/14 + 1 1/14

r = 1/7 1 + (3/14) 0 + 0 =

1/7

= 2/14 .

0

1

0

3/14

3/14

Hence, the "row space" solution reads

-2

-3

1

xr = 1 + 2 + r =

1

1

+

2

0

+ (1/14) 2 .

0

1

3

-2

-3

1

By

definition

1

1

+

2

0

N (A)

and

(1/14)

2

RA

0

1

3

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