Linear Algebra - 3

Hi,

Now that we have seen what span of a set of vectors is, let us see some more concepts. 

What is linear dependence of vectors?

Vectors are linearly dependent iff one vectors can be obtained from another(one vector is the scalar product of the other). For example, (3, 0) can be obtained from (1, 0). Thus, these two vectors are linearly dependent. There are different ways to check for linear dependency of vectors.

One way is as follows: 

Consider a linear combination of vectors a and b as follows:

$$ c_1 * \vec a+ c_2 * \vec b  = \vec 0 $$  

Now, if c1 or c2 is non-zero, the a and b are linearly dependent.

Linear Subspace of Rn

The linear subspace of Rn is defined as follows:

S is the linear subspace of Rn iff:

1. S contains the 0 vector.
2. For any vector X in S, c * X is also in S. (Closure under scalar multiplication).
3. For any 2 vectors a and b, (a + b) is in S. (Closure under addition).

Thus, if any set S in Rn satisfies the above 3 conditions, the it is a linear subspace of Rn.

Let us consider an example:

Is 0 vector a valid linear subspace under Rn?

Now, let us verify the 3 conditions:

- Does it have the 0 vector? yes
- Is it closed under scalar addition?
      Let us check: c * 0vec = 0vec. Thus c * 0vec belongs to the set. So, yes
- Is it closed under addition?
      0vec + 0vec = 0vec. Thus, it is closed under addition. So, yes

Thus, 0vector is a valid linear subspace under Rn.

Basis of a (linear) Subspace:

Consider a set:$$V=span( V_1 , V_2, V_3 ... V_n  |  V_i \in R^n)$$
Now, if this set is linearly independent, then (V1, V2, .. Vn) is a basis for V. 

So, what this means is, we can obtain any vector in the span of (V1, V2, .. Vn), by using a linear combination of the vectors (V1, V2, .. Vn).
Also, note that the basis should be a minimum set such that it covers the subspace.

Let us take an example and see this: Now, let $$ S = span( (1, 0) , (0, 1) ) $$
Clearly,  (1, 0) , (0, 1) are linearly independent. So, this can be a basis for span((1, 0) , (0, 1)). Now, let us take another example:


$$ S = span( (1, 0) , (0, 1), (4, 0)) $$
 Here,  (1, 0) , (0, 1), (4, 0) are not linearly independent since one of the vectors here is redundant(in that one can be generated from the other). Although
span((1, 0) , (0, 1), (4, 0)) would be a valid subspace, this is not the minimum set with which we can cover the subspace. Thus, in this case, S cannot be the basis for the span((1, 0) , (0, 1), (4, 0)).

Note: It is quite interesting to note that although (1, 0) and (0, 1) are chosen as the most common basis vectors for R2, there can be several other vectors that can be the basis for Rn!