Difference between revisions of "Linear Algebra the Conceptual Way" - Rhea

Revision as of 00:43, 6 November 2017

Introduction

For many students they are able to skate by in linear algebra by following equations and systems but don't understand the intuitive nature of matrices and vectors and their operators. This tutorial is not meant as a replacement to the course but should rather be used as a supplement to the course to understand why the operations work as they do. This tutorial is intended to be receivable by a wide range of individuals including past linear algebra students wanting review, present students seeking help, and my friends that I inevitable force to read my work.

Contents

1. Vectors
   1. Addition/Subtraction
   2. Scalar Multiplication
   3. Unit
   4. Dot
   5. Cross

Vectors

For computer science students vectors can be seen as ordered lists, for engineering students focused on physics they can be seen as a direction and a length. For linear algebra they can be approached from any and every angle Given $ 0\leq\theta<2\pi $ of course.

For the purposes of this tutorial think of it as a way to move a point (normally at the origin) to another point

As a warning most of this page will be movement oriented and I will try my best to graphically demonstrate that

So for example the vector written $ \begin{bmatrix} 1\\ 2\end{bmatrix} $ will move a vector from the origin to point (1,2)

If we want vectors to have all the properties of numbers then what should a vector + a vector result in.

What if we make it one movement and then the other? This way $ \begin{bmatrix} 1\\ 2\end{bmatrix}+\begin{bmatrix} -3\\ 2\end{bmatrix} $ will be the resultof moving right 1 and up 2 followed by moving left 3 and up 2.

As displayed by the animation this is the same as adding the x component each vector and the y component of each vector.

In this way $ \begin{bmatrix} 1\\ 2\end{bmatrix}+\begin{bmatrix} -3\\ 2\end{bmatrix}=\begin{bmatrix}1-3\\2+2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix} $

Vector subtraction is built on primarily the same process but in reverse as could be expected.

Some ways to derive the exact method would be imagining $ \vec{u}-\vec{v}=\vec{w} $ as $ \vec{u}=\vec{w}+\vec{v} $

We know from before that $ \begin{bmatrix} 1\\ 2\end{bmatrix}+\begin{bmatrix} -3\\ 2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix} $ So in that case $ \begin{bmatrix} 1\\ 2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}-\begin{bmatrix} -3\\ 2\end{bmatrix} $ or $ \begin{bmatrix}-2\\4\end{bmatrix}-\begin{bmatrix} -3\\ 2\end{bmatrix}=\begin{bmatrix} -2-(-3)\\ 4-2\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix} $

So vector subtraction works very much the way that we would expect as well.

You may be thinking that $ \begin{bmatrix} 1\\ 2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}-\begin{bmatrix} -3\\ 2\end{bmatrix} $ could also be proven by writing it as $ \begin{bmatrix} 1\\ 2\end{bmatrix}=\begin{bmatrix}-2\\4\end{bmatrix}+(-1)*\begin{bmatrix} -3\\ 2\end{bmatrix} $.

We haven't defined scalar multiplication yet but imagine it as adding a vector to itself so $ 2*\begin{bmatrix} 1\\2\end{bmatrix} $ is the same as $ \begin{bmatrix} 1\\2\end{bmatrix}+\begin{bmatrix} 1\\2\end{bmatrix} $

This would just be $ \begin{bmatrix} 1+1\\2+2\end{bmatrix} $ or $ \begin{bmatrix} 2*1\\2*2\end{bmatrix} $

So multiplying a vector by a scalar is the same as multiplying the individual components of the vector by that number

Unit Vectors

Since we seem to be using this specific example vector a lot, why don't we combine these two concepts to start writing $ \begin{bmatrix}1\\2\end{bmatrix} $ as $ \begin{bmatrix}1\\0\end{bmatrix}+2*\begin{bmatrix}0\\1\end{bmatrix} $?

The two vectors $ \begin{bmatrix}1\\0\end{bmatrix} $ and $ \begin{bmatrix}0\\1\end{bmatrix} $ seem like we'll use them a lot so lets name them $ \vec{i} $ and $ \vec{j} $ respectively.

Now instead of me typing out all the code required to show $ \begin{bmatrix}1\\2\end{bmatrix} $ $ \textsubscript{Which is a lot actually I don't know why I even bother} $, I can just say $ \vec{i}+2\vec{j} $.