Linear Independence in vector space
Definition.
The vectors v1,
v2, ..., v3 in the vector v space are called linearly independent
if
c1v1
+ c2v2 + ... +
cnvn = 0
resulting in all scalars c1, c2, ..., c3 must
be equal to 0.
Example
1.
The
vectors 
and 
are linearly independent, because if
Then,
c1 + c2 = 0
c1 + 2c2
= 0
And the only solution to
this system is c1 = 0, c2 = 0.
Definition.
Vectors v1, v2,
..., vn in vector v space
are called linearly dependent if there are scalars c1, c2,
..., cn which are not all zero so
c1v1
+ c2v2 + ... + cnvn = 0
Example 2.
Determine what vectors are
v1 = (1, -2,
3) v2 = (5, 6, -1) v3 = (3, 2, 1)
linear Independent or linearly
dependent?
Solution:
On the vector component
segment
c1v1
+ c2v2 + c3v3 = 0
to be
c1= (1, -2, 3)
+ c2(5, 6, -1) + c3(3,
2, 1) = (0, 0, 0)
or equivalent to
(c1 + 5c2
+ 3c3, -2c1 + 6c2 + 2c3, 3c1
– c2 + c3) = (0, 0, 0)
By equalizing the corresponding
component will give
c1 + 5c2
+ 3c3 = 0
2c1 + 6c2
+ 2c3 = 0
3c1 – c2
+ c3 = 0
By completing this system
it will produce
c1= -1/2t c2 = -1/2t c3 = t
So, this system has
non-trivial solutions, so the three vectors are linearly dependent.
Theorem 1.
Suppose x1, x2,
..., xn are n vectors in
Rn and for example
x1 = (x1i, x2i, ..., xni)T
untuk i = 1, 2, ..., n
if X = (x1, x2,
..., xn) then vectors x1, x2, ..., xn
are linearly dependent if and only if X is singular.
Proof:
The equation c1x1
+ c2x2 + ... + cnxn = 0 is
equivalent to the system of equations
If we say c = (c1,
c2, ..., cn)T, then this system can be written
as a matrix equation.
Xc = 0
This equation will have a
non-trivial solution if and only if X
is singular. So x1, x2, ..., xn will depend
linearly if and only if X is singular.
Example 3.
Determine whether vectors (4,
2, 3)T, (2, 3, 1)T
and (2, -5. 3)T are linearly dependent or not.
Solution:
Because
Then these vectors are
linear dependent.
SUBSCRIBE TO OUR NEWSLETTER
0 Response to "Linear Independence in vector space"
Post a Comment