### Linear Indpendence

##### 3. Properties of linear independence

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

v1, v2, ..., vk, vk+1 linearly independent ⇒ v1, v2, ..., vk linearly independent.

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.

v1, v2, ..., vk linearly independent and vk+1span{v1, v2, ..., vk} ⇒ v1, v2, ..., vk, vk+1 linearly independent.

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 cos2t and sin2t are linearly independent. Then we have

cos2t, sin2t are linearly independent
⇒ cos2t, 1 are linearly independent (1 = cos2t + sin2t)
⇒ -2cos2t, 1 are linearly independent
⇒ cos2t, 1 are linearly independent (cos2t = 1 - 2cos2t)
⇒ cos2t, 2sin2t are linearly independent (2sin2t = 1 - cos2t)
⇒ cos2t, sin2t are linearly independent.

Example In this example, we effectively showed that the vectors v1 = (1, -1, 0, 1), v2 = (3, -1, 4, 3), v3 = (2, -1, 2, 2), v4 = (0, 1, 4, -2), v5 = (1, 0, 3, 0) are linearly independent by row operations on the matrix

 A = [v1 v2 v3 v4 v5] = [ 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 v1, v2, v3, v4, v5 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 AB by column operations, then columns of A are linearly independent ⇔ columns of B are linearly independent.

Second is the numerical consequence of linear independence.

v1, v2, ..., vkRn are linearly independent ⇒ kn.

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