The first property concerns how linear independence is changed when we add/delete vectors (compare with this property of span).

`v`_{1}, `v`_{2}, ..., **v**_{k}` _{}`,

It is easy to understand the property from the intuition that linear independence means no waste: If all five people are necessary for the job, then surely any four of the five (or any three, any two, etc) are also necessary.

The following is the converse of the property above.

`v`_{1}, `v`_{2}, ..., **v**_{k}` _{}` linearly independent and

Similar to span, the following result is useful for computing linear independence.

The three operations do not change linear independence.

The proof of the two properties can be found here.

Example By the method similar to this example, we know
cos^{2}`t` and sin^{2}`t` are linearly independent. Then we have

cos^{2}`t`, sin^{2}`t` are linearly independent

⇒ cos^{2}`t`, 1 are linearly independent (1 = cos^{2}`t` + sin^{2}`t`)

⇒ -2cos^{2}`t`, 1 are linearly independent

⇒ cos2`t`, 1 are linearly independent (cos2`t` = 1 - 2cos^{2}`t`)

⇒ cos2`t`, 2sin^{2}`t` are linearly independent (2sin^{2}`t` = 1 - cos2`t`)

⇒ cos2`t`, sin^{2}`t` are linearly independent.

Example In this example, we effectively showed that the vectors
`v`_{1} = (1, -1, 0, 1),
`v`_{2} = (3, -1, 4, 3),
`v`_{3} = (2, -1, 2, 2),
`v`_{4} = (0, 1, 4, -2),
`v`_{5} = (1, 0, 3, 0) are linearly independent by
*row operations* on the matrix

= [Av_{1} v_{2}
v_{3} v_{4} v_{5}] =
[ |
1 | 3 | 2 | 0 | 1 | ]. |

-1 | -1 | -1 | 1 | 0 | ||

0 | 4 | 2 | 4 | 3 | ||

1 | 3 | 2 | -2 | 0 |

Alternatively, we may use the *column operations* on ** A** to simplify the vectors and get the answer. By the computation in the earlier example, the column echelon form is

[ | 1 | 0 | 0 | 0 | 0 | ]. |

-1 | 1 | 0 | 0 | 0 | ||

0 | 2 | 1 | 0 | 0 | ||

1 | 0 | -1 | 0 | 0 |

The five column vectors in the column echelon form are clearly linearly dependent
(for example, the linear combination with coefficients 0, 0, 0, 1, 0 is the zero vector).
Since linear independence is not changed by the three operations, we conclude that
`v`_{1}, `v`_{2}, `v`_{3},
`v`_{4}, `v`_{5} are linearly dependent.

The example above leads to some interesting observations. First for euclidean vectors, the preservation of the linear independence under the three operations can be rephrased in terms of column operations.

If ** A** →

Second is the numerical consequence of linear independence.

The equivalent statement in terms of the null space appeared in the first part of this exercise.

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[Extra: Properties of linear independence]