The set {v_{1}, v_{2}, v_{3}, ............, v_{n}} is linearly dependent if the equation
b_{1}v_{1 }+ b_{2}v_{2 }+ b_{3}v_{3 }+ ............ + b_{n}v_{n} = 0
such that
b_{1}_{ }or b_{2}_{ }or b_{3}_{ }............ or b_{n} ≠ 0
That is, all the weights b_{1}, b_{2}, b_{3}_{ }............ b_{n} are not zero.
A collection of vectors {v_{1}, v_{2}, v_{3}, ............, v_{n}} is linearly independent if the equation
a_{1}v_{1 }+ a_{2}v_{2 }+ a_{3}v_{3 }+ ............ + a_{n}v_{n} = 0
implies that
a_{1}_{ }= a_{2}_{ }= a_{3}_{ }= ............ = a_{n} = 0
It has only trivial solution.
Example 1 :
Show the following set of vectors is linearly dependent.
{(1, 2, 1), (0, 1, 0), (-2, 0, -2)}
Solution :
Let a, b and c be the scalars.
We can write
2b - (½)c = 2(0, 1, 0) - (½)(-2, 0, -2)
2b - (½)c = (0, 2, 0) - (-1, 0, -1)
2b - (½)c = (0, 2, 0) + (1, 0, 1)
2b - (½)c = (1, 2, 1)
2b - (½)c = a
Since a can be written as a linear combination of b, c, the set is linearly dependent.
Example 2 :
Show the following set of vectors is linearly independent.
{(1, 0, 1), (1, 1, -1), (0, 1, 0)}
Solution :
Let a, b and c be the scalars. Then, we have
a(1, 0, 1) + b(1, 1, -1) + c(0, 1, 0) = (a + b, b + c, a - b)
The zero vector is (0, 0, 0).
If the given set of vetcors is linearly independent,
a(1, 0, 1) + b(1, 1, -1) + c(0, 1, 0) = 0
Then, we have
a + b = 0 ----(1)
b + c = 0 ----(2)
a - b = 0 ----(3)
From (3), we see that a = b.
Substitute b = a into (1).
a + a = 0
2a = 0
a = 0
Since, a = b, we have
b = 0
Substitute b = 0 into (2).
0 + c = 0
c = 0
Since all of the scalars a, b and c are zero, the set is linearly independent.
We can show that a set of vectors is linearly independent by arranging them in a matrix form. Then row reduce the matrix; if each row has a nonzero pivot, then the vectors are linearly independent.
Example 3 :
Determine if the following set of vectors is linearly independent.
{(1, 3, 5) , (4,−1, 2) , (0,−1, 2)}
Solution :
We can arrange the given vectors as a matric, using each vector as a column. Then, the matrix is
Now we row reduce the matrix.
Since all the columns in the reduced matrix contain a pivot entry, no vector can be written as a linear combination of the other vectors; therefore, the set is linearly independent.
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