MATA23H3 Lecture Notes - Lecture 12: Row And Column Spaces, Gaussian Elimination, A127 Road

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MATA23 - Lecture 12 - Bases, Dimensions, Spaces, and Nullity
Bases Continued
Example 5:
1
1
0
,
0
1
0
,
1
1
0
,
0
1
0
,
1
0
0
W1=sp(
1
1
0
,
0
1
0
) =
r
1
1
0
+s
0
1
0
|r, s R
is a subspace of R3
Since
(
1
1
0
,(
0
1
0
is linearly independent
r
1
1
0
+s
0
1
0
=
0
0
0
=r=s= 0
1
1
0
,
0
1
0
is a basis of W1by theorem
W2=sp(
1
1
0
,
0
1
0
,
1
0
0
), a subspace of R3
=
r1
1
1
0
+r2
0
1
0
+r3
1
0
0
|riR, i = 1,2,3
Note that r1
1
1
0
+r2
0
1
0
+r3
1
0
0
=
0
0
0
101
110
000
101
011
000
1
1
0
,
0
1
0
,
1
0
0
is linearly independent (not a basis for W2)
~
0 =
0
0
0
= 1 ·
1
1
0
+ (1) ·
0
1
0
+ (1) ·
1
0
0
~
0=0·
1
1
0
+ 0 ·
0
1
0
+ 0 ·
1
0
0
; not unique
Let Wbe a subspace of Rn.n~
b1,~
b2,...,~
bkoWis a basis for Wif and only if:
1
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1. ~
b1,~
b2,...,~
bkspan W=W=sp(~
b1,~
b2,...,~
bk)
2. ~
b1,~
b2,...,~
bkare linearly independent =r1
~
b1, r2
~
b2, . . . , rk~
bk=~
0
ri= 0; i= 1,2, . . . , k
To find a basis for W=sp(~v1, ~v2, . . . , ~vk):
1. Form the matrix Awhose jth column vector is ~vj
2. Row reduce Ato row-echelon form H
3. The set of all ~vjsuch that the jth column of Hcontains a pivot is a basis for W
Example 6: Let ~v1=
1
1
0
2
1
, ~v2=
2
1
2
0
0
, ~v3=
0
3
2
4
2
, ~v4=
3
3
4
2
1
, ~v5=
2
4
1
0
1
, ~v6=
5
7
3
2
0
.
Find a basis for W=sp(~v1, ~v2, ~v3, ~v4, ~v5, ~v6)
1 2 0 3 2 5
1 1 3347
02 2 4 1 3
2042 0 2
1 0 2 1 1 0
R2R2+R1
R4R42R1
R5R5R1:
1 2 0 3 2 5
0 3 3 6 6 12
02 2 4 1 3
04 4 8412
02 2 215
R21
3R2:
1 2 0 3 2 5
0 1 1 2 2 4
02 2 4 1 3
04 4 8412
02 2 215
R3R3+ 2R2
R4R4+ 4R2
R5R5+ 2R2:
1 2 0 3 2 5
0 1 1 2 2 4
0 0 0 0 5 5
0 0 0 0 4 4
0 0 0 2 3 3
2
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R3R5
R41
4R4:
1 2 0 3 2 5
0 1 1 2 2 4
0 0 0 2 3 3
0 0 0 0 1 1
0 0 0 0 5 5
R5R55R4:
1 2 0 3 2 5
0 1 1 2 2 4
0 0 0 2 3 3
0 0 0 0 1 1
0 0 0 0 0 0
A basis for Wis {~v1, ~v2, ~v4, ~v5}
Example 7: Find a basis for the solution space of the homogenous system
x1x2+x3= 0
x1+ 2x2+ 2x3x4= 0
x1+ 8x3x4= 0
x1+ 4x2+ 12x33x4= 0
11 1 0 |0
1 2 2 1|0
1 0 8 1|0
1 4 12 3|0
R2R2+R1
R3R3R1
R4R4+R1:
11 1 0 |0
0 1 3 1|0
0 1 7 1|0
0 3 13 3|0
R3R3R2
R4R43R2:
11 1 0 |0
0131|0
0040|0
0040|0
. . .
11 1 0 |0
0131|0
0010|0
0000|0
x3= 0
x4= 5
3
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