Important Example Solved Problems - Engineering Maths or Mathematics - Bases and Dimensions
BASES AND
DIMENSIONS
WORKED
EXAMPLES
Example 1
In Rn, let e1
= (1, 0, 0, ... 0), e2 = (0, 1, 0, 0, ...0)…., en = (0,
0, ...1).
Prove that the set S={e1,
e2, ... en} is basis of Rn.
Solution:
Let
a1e1 + a2e2 + ... + anen
= 0
a1(1,
0, 0, 0... 0) + a2 (0, 1, 0, 0, ... 0) + ... + an(0,
0, 0, ... 1) = 0
(a1, 0, 0, ... 0) + (0, a2, 0... 0) + ... + (0, 0, 0, ... an)
=
(0,0...0) (a1, a2, ... an) = (0, 0, ... 0)
a1
=0, a2 =0,... an = 0
S
is linearly independent.
Let
(x1, x2, ... xn)
∈ Rn
(x1,
x2,... xn) = x1
(1, 0, ... 0) + x2 (0, 1, 0, ... 0) + ... + xn(0, 0, ...
1)
L(S)
= V
S is a Basis of V.
Example 2
Let S = {v1,
v2, v3 } where v1 = (2, 1, 0), v2 =
(−3, −3, 1) & v3 = (− 2, 1, − 1). Show that S is a basis of R3.
Solution:
To
check these vectors are linearly independent
= 2(3−1) − 1 (3 + 2) = 4 − 5 =−1 ≠ 0
The
vectors are linearly independent.
Let
(x, y, z) ∈
R3
(x, y, z) = a(2, 1, 0) + b(−3, −3, 1) + c(− 2,
1, − 1)
(x,
y, z) = (2a, a, 0) + (−3b, − 3b, b) + (−2c, c, −c)
2a−3b−2c
= x ……(1)
a−3b+c=y ……(1)
b−c=z
……(1)

−c=x−2y−3z
c
= 2y+3z−x
b=−x+2y+4z
a=−2x+5y+9z
L(s) = V
S
generates V.
S
is a basis.
Example 3
In Mm×n (F),
let Eij denote the matrix whose only non−zero entry is a 1 in the ith
row and jth column. Then prove that { Eij,1≤i≤m,1≤j≤n}
is a basis for Mm×n (F).
Solution:
Let
V = M2×3 (F)

Note:
1.
In Pn (F) the set {1, x, x2, ... xn) is a
basis of degree n.
2.
In P(F) the set {1, x, x2,...} is a basis.
A
basis need not be finite. Not every vector space has a finite basis.
Example 4
Prove that every
non−zero singleton set is linearly independent for V=R.
Solution:
S={0}
is linearly dependent.
α.0=0
need not be a=0
α ≠ 0
S is linearly dependent.
S={}
is linearly independent.
α.ϕ=0
α=0
S is linearly independent & it has a
dimension zero.
Linear Algebra: UNIT I: Vector Spaces : Tag: maths, mathematics : - Bases and Dimensions: Example Solved Problems
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