Linearly independent vectors do not lie on the same line in a coordinate system. Am important application of this is that you can use linearly independent vectors to form a basis for a vector space. Let me explain what that means.

Consider the following vectors:

(1) and (0) (0) (1)

As you can see, these two vectors do not lie on the same line: one is going in, say, the +x direction and the other in the +y direction. Therefore, these two vectors are linearly independent.

Another interesting observation about these is that they span R2, the two-dimensional vector space of real numbers. Think about R2 as all of the points you can choose on a Cartesian plane. These vectors span R2 because by multiplying each of them by a different weight you can produce any vector in R2, including the zero vector. These linear combinations look like this:

(a) = c(1) + d(0) (b) (0) (1)

You could have a set that spans, but is not linearly independent: add

(2) (0)

to that set.

Or, you could have a set that is linearly independent, but does not span: just the vector

(1) (0)

for example.

Because our set is a linearly independent spanning set, we call it a basis for R2. A basis is an efficient spanning set. You can see that a basis must be linearly independent to span with the fewest vector possible.

There are other applications of linear independence, but the ability to create an efficient set that spans all vectors in a given vector space is responsible for many of them.

Consider facial recognition for example. Linear algebra is used in facial recognition software, so this is a very simplified form. Imagine you wanted to be able to make any face. You'd need eyes, a nose, a mouth, etc., which can be represented by linearly independent vectors (sorry, this is where I'm glossing over a lot of stuff -- hopefully someone else has a better knowledge of this than I). By creating efficient sets of those linearly independent facial features, you can span the features of a particular face. (This has the most to do with eigenvalues and eigenvectors, but that's a different beast.)

Source: linear algebra class