Linear Algebra - 2

Hi,

In the last post, we say some basics of vectors. Now, let us dive a little more into the topic.  Now, we saw that vectors can be added, subtracted and multiplied(cross and dot). Vectors can also be multiplied by constants.

So,$$c*V=\{ c*X_1 , c* X_2,  c*X_3 ... c*X_n  |  c_i \in R \}$$ 

Now, a linear combination of entities is an expression obtained by multiplying each entity by a constant and adding results.

eg: ax + by(where a and b are constants) is a linear combination of x and y.

What is a linear combination of vectors?

A linear combination of vectors is defined as follows:

$$\{ c_1*X_1  + c_2 *  X_2 + c_3 * X_3 + ... c_n * X_n  |   c_i \in R \}$$ 

This represents all the  linear combinations of vectors (X1...Xn).

Span of a set of vector(s)

Now that we saw what it means by a linear combination of vectors, now let us define an interesting concept called as span. The span of a set of vectors is defined as follows:

$$span(X_1, X_2, ... X_n)=\{ c_1*X_1  + c_2 *  X_2 + c_n * X_3 + ... c_n * X_n \}$$ 

Thus, the span of a set of vectors is the set of all linear combination of the vectors.

Now, if we take the example. $$span( (1, 0), (0, 1) )$$
So, by definition, span is the set of all the linear combination of (1, 0) and (0, 1).  Now, we can make an interesting question.

Can the span of vectors constitute Rn?

What this mean is, can the span cover the entire space? Now, let us see our above example of (1, 0) and (0, 1). Clearly, the span gives the following:

$$span( (1, 0), (0, 1) )=\{ c_1 * (1, 0)  + c_2 *  (0, 1) + c_3 * (1, 0) + c_4 * (0, 1) + ... \}$$

So, we can have all sorts of constants, any number of times. Now, let us see if this covers the entire two dimensional space R2. Can we derive an arbitrary vector (x, y) 
from these two vectors? Let us see.

Now, if we need to derive any vector (x, y) from the above two vectors, then we have
$$ c_1 * (1, 0) + .. +  c_m * (1, 0) + c_2 * (0, 1) + ... + c_n * (0, 1) = (x, y) $$
$$ c_1 + ... + c_m = x $$
$$ c_2 + ... + c_n  = y $$

Now, since all of the constants "c"s belong to R, it is definitely possible to obtain a solution for x and y, as long as there is at least one non-zero constant. Thus, if we obtain a non-zero solution for the equation above, this means that the span( (1, 0), (0, 1) ) indeed covers R2.

Can span() of any vectors constitute Rn?

No, the span() would constitute Rn iff every vector of Rn can be obtained by the vectors. Now, clearly, span((1, 0), (0, 1)) constitutes R2. However, take the following example: $$span((1, 0), (2, 0))$$

Clearly, any linear combination of (1, 0) and (2, 0) cannot generate for example, the vector (1, 1). Thus, the span of these two vectors cannot constitute Rn.