Linear combinations, linear independence, span

Read LA 1.3, 2.0 - 2.3, 3.0 - 3.3


1) Main Points

Example 1.3 reminds us that we can combine three vectors (or more) by using the parallelogram or 'tip to tail' method to find the resultant. Also, by using the concept of linear combination we can combine multiples of vectors. Chapter 2 continues with this concept and asks how to find linear a combination that equals a different desired vector. We can do this by using x and y as variables in an equation. If we have the vectors (2,3),(-1,1) and (0,5) and we want the first two together to equal the third we can write it like this:

x(2,3) + y(-1,1) = (0,5)

At this step it is possible to test solutions for x and y but a more efficient way is to solve for an answer like so:

The previous equation shown above is equal to:

(2x,3x) + (-y,y) = (0,5)

Because vectors add across (remember they are usually written vertically) this is equal to:

(2x-y,3x+y) = (0,5)

This gives us a system of equations like so:

2x-y=0
3x+y=5

This can be solved which gives us x=1, y=2 as our solution to the problem above. This means that if we multiply the first vector by 1 and add it to the second vector multiplied by 2 it will be equal to the third.

It is also possible, using algebra to go backwards from a system of equations to three vectors so make sure you know how to do this. A problem with this method however is that sometimes two vectors will be in the same direction and therefor can't be expressed by using a system of equations because the lines may not intersect. In this case they are most easily thought of by thinking of what variables give the desired vector because the linear set up is easier to think about. Creating a system of equations to find a solution is applicable in dimensions beyond 2, however they will produce a system of however many dimensions there are. Lists of vectors can be listed as matrices. This allows us to rewrite vector equations as matrices:

For example:

x(2,3) + y(-1,1) = (0,5)

can be

(2,3, -1,1)(x,y) = (0,5)

(this doesn't come across well because I can't write it vertically, the point is that matrices can consolidate the equation). This means Ax = b, A is the matrix of vectors, x is the variables matrix and b is the goal.

Chapter 3 introduces a 'span" which is a list of all the linear combinations that can be made from the given vectors.

(6,2) is the span of vectors (2,1),(1,1) because 4(2,1) -2(1,1) = (6,2)

A list of vectors is linearly independent if no vector on the list is a linear combination of the other vectors. Otherwise the list is linearly dependent.

2) Challenges

I followed these examples pretty well until the concept of a "span". I don't understand span and I think I need to know more about what a list of linear combinations looks like. Other than that I can see some difficulties in writing vectors with matrices in mind. This to me seems like it could be a little confusing. Also it seems a little difficult to go backwards from a system of equations, this is something I should practice.

3) Reflections

I remember using vectors a lot in physics, they can really account for forces acting in different (or the same) directions. I think linear algebra will seem more applicable when we learn about where to actually use it.