ORIE 3300/5300  Fall 2010
Solutions for Recitation 2
1. Suppose
y
and
z
are two solutions of
Ax
=
b
, that is,
Ay
=
b
and
Az
=
b
. Hence, we
get
A
(
y

z
) = 0. Let
u
:=
y

z
.
By substitution, we have
Au
= 0.
So, if
A
1
, A
2
, ...A
n
are the columns of
A
, then
∑
n
j
=1
u
j
A
j
= 0.
The linear independence of the columns of
A
implies that
u
j
= 0
∀
j
= 1
, ...n
. This implies that
y
=
z
. Hence, there is only one solution of
Ax
=
b
.
2.
(a) The column is linearly independent.
The column does not span (not enough columns to span).
There is no basis.
The matrix is not invertible (only a square matrix can be invertible).
(b) The columns are linearly independent.
The columns span.
There are two bases: [1,2] and [2,1]
The matrix is invertible.
(c) The columns are not linearly independent: we have 2
1

3
+

2
6
=
0
0
The columns do not span.
There is no basis.
The matrix is not invertible (because the columns are not linearly independent).
(d) The columns are linearly dependent (there are too many columns to be linearly
independent, and the same relationship as in (c) holds)
The columns span.
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 Fall '08
 TODD
 ERO

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