Is the case u 1 u m is said to be linearly dependent

is the case, { ˛ u
1
, . . . , ˛ u
m
} is said to be linearly dependent .

10 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES)
We want to focus on sets { ˛ u
1
, . . . , ˛ u
m
} that have no such redundancy. Such a set of
vectors is called linearly independent ; formally, this means that the only set of scalars
–
i
, i = 1 : m such that
m
ÿ
n =1
–
n
˛ u
n
=
˛
0
is –
i
= 0, for all i .
Definition 5. A set { ˛ u
1
, . . . , ˛ u
m
} of vectors in W forms a basis for W i ff these two
properties are true:
1. span ( ˛ u
1
, . . . , ˛ u
m
) = W
2. { ˛ u
1
, . . . , ˛ u
m
} is linearly independent .
In other words, “ { ˛ u
1
, . . . , ˛ u
m
} is a basis for W ” means ˛ x =
q
m
n =1
–
n
˛ u
n
is uniquely
solvable for any ˛ x œ W .
Examples 3. 1. Standard basis for both for R 3 and C 3 :
e
1
= (1, 0, 0)
T
, e
2
= (0, 1, 0)
T
, e
3
= (0, 0, 1)
T
.
Check this is a basis: For any X = ( x
1
, x
2
, x
3
)
T
, X =
q
i
–
i
e
i
is uniquely solved by
x
i
= –
i
, i = 1, 2, 3 . Note that in case V = C 3 (respectively R 3 ), the scalars are
complex (real).
2. u
1
= (1, 1, 0) T , u
2
= (1, ≠ 1, 0) T , u
3
= (1, 1, 1) T is another basis for R 3 and C 3 . Why?
Answer by row reduction: For any X , to find A = ( –
1
, –
2
, –
3
)
T
such that
X = ( x
1
, x
2
, x
3
)
T
= –
1
u
1
+ –
2
u
2
+ –
3
u
3
we use the same trick as before. We write U = [ u
1
, u
2
, u
3
] and solve
X = UA
for the 3 unknown components of A . By row reducing the augmented matrix [ U | X ] ,
we can show the system is uniquely solvable for any X .
More succinctly, for any X œ C
3
, its coordinates A œ C
3
are computed by A = U
≠ 1
X
where U ≠ 1 is the inverse of U .
Algorithm 1. (Computing U ≠ 1 ): use Gauss-Jordan elimination on the fully aug-
mented [3, 6] matrix
[ U | I ] æ [ I | U
≠ 1
]
Exercise: Review Algorithm 1 and compute the inverse of U =
Q
c
a
1 1 1
1 ≠ 1 1
0 0 1
R
d
b
.

1.4. LINEAR OPERATORS 11
3. V = C [0, 1]: The three functions { cos x , sin x , 1 } are linearly independent, which
means they form a basis for their span. On the contrary, the three functions
{ cos
2
x , sin
2
x , 1 } are linearly dependent because of the identity
1 · cos
2
x + 1 · sin
2
x + ( ≠ 1) · 1 = 0 (1.5)
4. The functions v
n
( x ) = e inx for n = ≠ N : N form a basis for the vector space W N .
5. The pair of functions J
n
, Y
n
form a basis for the solution space of (1.1). We some-
times say they form a fundamental set of solutions .
Definition 6. 1. V has dimension N means it has a set of N linearly independent
vectors, but no set of N + 1 linearly independent vectors
2. V is infinite dimensional means it has a set of N linearly independent vectors for
every N .
Theorem 1. (“Dimension theorem”) Any two bases { ˛ u
1
, . . . , ˛ u
m
} and { ˛ v
1
, . . . , ˛ v
n
} for a
vector space V with finite dimension N have the same number of vectors m = n = N .
Examples 4. 1. R
3
has dimension 3 . So does C
3
.
2. V = C [0, 1] is infinite dimensional because one can show that { 1, x , x 2 , . . . , x m } is
linearly independent for all m . (Homework!)
3. The solution space of (1.1) is two-dimensional.
Given a specific basis { ˛ u
n
} N
n =1
of an N -dimensional vector space V , every ˛ x œ V can
be uniquely expressed in the form
˛ x =
N
ÿ
n =1
–
n
˛ u
n
The components of a = ( –
1
, . . . , –
N
) T form a column N -vector a œ C N is called the
coordinate N -vector of ˛ x relative to the basis { ˛ u } , or the u