Linearly dependent and linearly independent vectors
Definition. A linear combination of vectors a_{1}, ..., a_{n} with coefficients x_{1}, ..., x_{n} is a vector
x_{1}a_{1} + ... + x_{n}a_{n}.
That is, the vector a_{1}, ..., a_{n} are linearly independent if x_{1}a_{1} + ... + x_{n}a_{n} = 0 if and only if x_{1} = 0, ..., x_{n} = 0.
Linearly dependent vectors properties:
- For 2-D and 3-D vectors.Two linearly dependent vectors are collinear. (Collinear vectors are linearly dependent.)
- For 3-D vectors.Three linear dependence vectors are coplanar. (Three coplanar vectors are linearly dependent.)
- For an n-dimensional vectors.n + 1 vectors always linearly dependent.
Linearly dependent and linearly independent vectors examples:
Solution:
The vectors are linearly dependent, since the dimension of the vectors smaller than the number of vectors.
Solution: Calculate the coefficients in which a linear combination of these vectors is equal to the zero vector.
x_{1}a + x_{2}b + x_{3}c_{1} = 0This vector equation can be written as a system of linear equations
x_{1} + x_{2} = 0 | |
x_{1} + 2x_{2} - x_{3} = 0 | |
x_{1} + x_{3} = 0 |
Solve this system using the Gauss method
1 | 1 | 0 | 0 | ~ | ||
1 | 2 | -1 | 0 | |||
1 | 0 | 1 | 0 |
from 2 row we subtract the 1-th row;from 3 row we subtract the 1-th row:
~ | 1 | 1 | 0 | 0 | ~ | 1 | 1 | 0 | 0 | ~ | ||||
1 - 1 | 2 - 1 | -1 - 0 | 0 - 0 | 0 | 1 | -1 | 0 | |||||||
1 - 1 | 0 - 1 | 1 - 0 | 0 - 0 | 0 | -1 | 1 | 0 |
from 1 row we subtract the 2 row; for 3 row add 2 row:
~ | 1 - 0 | 1 - 1 | 0 - (-1) | 0 - 0 | ~ | 1 | 0 | 1 | 0 | ||||
0 | 1 | -1 | 0 | 0 | 1 | -1 | 0 | ||||||
0 + 0 | -1 + 1 | 1 + (-1) | 0 + 0 | 0 | 0 | 0 | 0 |
This solution shows that the system has many solutions, ie exist nonzero combination of numbers x_{1}, x_{2}, x_{3} such that the linear combination of a, b, c is equal to the zero vector, for example:
-a + b + c = 0
means vectors a, b, c are linearly dependent.
Answer: vectors a, b, c are linearly dependent.
Solution: Calculate the coefficients in which a linear combination of these vectors is equal to the zero vector.
x_{1}a + x_{2}b + x_{3}c_{1} = 0This vector equation can be written as a system of linear equations
x_{1} + x_{2} = 0 | |
x_{1} + 2x_{2} - x_{3} = 0 | |
x_{1} + 2x_{3} = 0 |
Solve this system using the Gauss method
1 | 1 | 0 | 0 | ~ | ||
1 | 2 | -1 | 0 | |||
1 | 0 | 2 | 0 |
from 2 row we subtract the 1-th row;from 3 row we subtract the 1-th row:
~ | 1 | 1 | 0 | 0 | ~ | 1 | 1 | 0 | 0 | ~ | ||||
1 - 1 | 2 - 1 | -1 - 0 | 0 - 0 | 0 | 1 | -1 | 0 | |||||||
1 - 1 | 0 - 1 | 2 - 0 | 0 - 0 | 0 | -1 | 2 | 0 |
from 1 row we subtract the 2 row; for 3 row add 2 row:
~ | 1 - 0 | 1 - 1 | 0 - (-1) | 0 - 0 | ~ | 1 | 0 | 1 | 0 | ~ | ||||
0 | 1 | -1 | 0 | 0 | 1 | -1 | 0 | |||||||
0 + 0 | -1 + 1 | 2 + (-1) | 0 + 0 | 0 | 0 | 1 | 0 |
from 1 row we subtract the 3 row; for 2 row add 3 row:
~ | 1 - 0 | 0 - 0 | 1 - 1 | 0 - 0 | ~ | 1 | 0 | 0 | 0 | ||||
0 + 0 | 1 + 0 | -1 + 1 | 0 + 0 | 0 | 1 | 0 | 0 | ||||||
0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
This means that the system has a unique solution x_{1} = 0, x_{2} = 0, x_{3} = 0, and the vectors a, b, c are linearly independent.
Answer: vectors a, b, c are linearly independent.
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