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Unformatted text preview: (1) First note that it suces to show Au = 0 A u = 0. For, if y and z are
two solutions of Ax = b, then A(y z ) = 0, showing, by the above,
that y z = 0 and hence y = z .
Now assume Au = 0. If a1 ; a2 ; : : : ; an are the columns of A, this means
Pn
j =1 uj aj = 0.
The linear independence of the columns now implies that uj = 0Vj ,
and thus completing the proof. (2) (i) The column does not span.
The column is linearly independent.
There is no basis.
The matrix is not invertible.
(ii) The columns span.
The columns are linearly independent.
There are two bases: f1; 2g and f2; 1g.
The matrix is invertible.
(iii) The columns do not span.
The columns are linearly dependent.
There is no basis.
The matrix is not invertible.
(iv) The columns span.
The columns are linearly dependent.
The matrix has four bases: f1; 2g, f1; 3g, f2; 1g and f3; 1g.
The matrix is not invertible. (3) Let us label the equations:
x1 x1 + x2 2x2 + (4) 2 ¢ (2) A x3 2x3
4x2 + 2x3 (1) (3) A
2 ¢ (5) (2) A x3 x2 = 2 (1)
+ x4 = 1 (2)
+ 2x4 = 7 (3)
+ 3x4 = 4 (4) + x3 2x4 = 5 (5) x3 5 x4 = 11 (6) x4 =2 1 Substituting this into (6) we get
x3 = 1 Substituting all the above into (3) we get:
x1 =1 x2 =0 Clearly (1) shows:
(4) Dividing the rst row by 2 gives 2
10
4 11 1
2
1
2 3 1
2 00
0 1 05 2 0 0 0 0 1 Adding 1 times the rst row to the second row gives: 3
2
1
101
2
200
4 0 1 0 1 1 0 5
2 2 00 001 Adding twice the rst row to the third row gives: 2
3
1
101
2
200
4 0 1 0 1 1 0 5
2
001 Finally, adding 101 1 times the third row to the rst row gives
2
2
13
100
0 0 2
1
4 0 1 0 2 1 0 5
001 10 1 This is clearly the required form with 3
2
0 0 1
2
1
05
C=4 21
10 1 Note that the initial matrix was of the form [DjI ] and whatever we
did above shows P [DjI ] = [I jC ] for some matrix P .
Unpacking this gives us:
PD
PI 2 =
= I
C Thus, D = P 1 = C 1 , showing 3
2
201 1 = 4 1 1 1 5
C
2 2 3 00 ...
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 '08
 TODD
 Linear Algebra, Linear Independence, Vector Space, basis, rst row

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