Linear algebra is easily the most interesting math class I’ve taken in my academic career. While the topics can be especially conceptual, the foundational logic has always seemed more straightforward to me than other subjects. I took a 200-level linear algebra class last year as a senior in high school and am now working on the 400-level here at Penn State. We haven’t really gotten to anything I haven’t done previously yet, so this week I’ll just talk about some fun matrice properties.
Firstly, we will be working with 2-dimensional vectors for simlicity, and to operate on these vectors there are two things we can do: add and multiply by a scalar.
When we add vectors we combine them tail to tip to make a new vector. In the example below, we add the vectors <1,2> and <3,1> to get the new vector <4,3>:
Very Cool.
Next, we can also multiply vectors by scalars to stretch and compress them. For example, if we take the orange vector <4,3>, we can multiply it by the scalar, 2, to get a new vector <8,6>.
Likewise, we can also shrink a vector by multiplying it by the scalar 1/2.
Armed with the knowledge of how to operate on vectors, we can begin to perform simple arithmetic with them. For example, if I have the vectors <5,-2> and <3,7> can I stretch and add them to make the vector <6,4>? If so, then what do we multiply each vector by to add them?
To find this out, we can do a lot of things because 2-dimensional vectors are a little silly, but the best thing to do is make a matrice.
We constructed this matrix by making our constituent vectors vertical and putting our answer vector in the last column [1]. With this setup, we can begin performing elementary row operations on the matrix to simplify it. The exact reason why we’re doing this and why it works is a bit out of the purview of this blog, so you’ll just have to trust me for now. Because this is only two rows, we only need to perform one operation, we take each element of row 1 and multiply it by 2, and then each element of row two and multiply them by 5, and then add each element in row one to its corresponding element in row two.
These operations leave us with a 0 in the first column and second row. This is good because it makes it much easier to tell what we need to multiply our vectors by to sum to <6,4>. We want our first two vector columns to be multiplied out to sum to the last value in their respective rows, so this is the same as solving a linear system which you might be more familiar with.
In this form, it should hopefully be easy to see that the only value for y is 32/41 because (41/41)*32 = 32. And if we know that y = 32/41 then it is easy to solve for x even though the numbers are ugly.
With this, we have our answer to the original question. If we multiply <5,-2> by 30/41 and <3,7> by 32/41 and then add them together, we get <6,4>. You can verify this for yourself if you’d like. Cool cool.
[1] – I feel like the vertical vectors could be a little confusing. They’re the same as writing <5,-2> and they multiply in the same way except they’re easier to work within the matrix.