User:Turiski

This guide to linear algebra is intended to aid a friend. For actual information about linear algebra, please see Linear Algebra

Vocab 0:
A set is a collection of mathematical objects. In the beginning they will be numbers, vectors, or matricies. Later they will be more abstract things.


Notation 0:
$a\in S$ is read "a exists in S," with the usual interpretation is that S is a set.

$\exists a\rightarrow P$ is read "there exists some (number/vector/etc) - called a - such that some property - called P - is satisfied"

$\forall a$ is read "for all [relevant] (numbers/vectors/etc)"

To combine the above two, as in "for all A there exists a B" or "there exists an A and a B," another $\rightarrow$ is needed

$\mathbb {N}$ is the set of natural numbers. $\mathbb {Z}$ for the integers, $\mathbb {Q}$ for the rationals, $\mathbb {R}$ for the reals, and $\mathbb {C}$ for complex numbers.

$\mathbb {R} ^{*}$ is the reals without 0, $\mathbb {Z} ^{+}$ is the positive integers. Similar notations apply for the others.


Example 0:
$\forall (n\in \mathbb {N} )\rightarrow \exists (x\in \mathbb {Z} )\rightarrow n+x=2$ reads "For all n in the natural numbers, there exists an x in the integers such that n+x = 2"

Or the more easily read "for a natural n, some integer x allows n+x = 2 to be true"


Vocab 1:
A set $S$ is "closed under * (star)" if $[\forall (a\in S)\rightarrow \forall (b\in S)]\rightarrow (a*b)\in S$. Note: * can be any operation that acts like addition or multiplication. In particular, + and x are examples of *s.

A set $V$ is a vector space (over the reals) if:

• It is closed under addition.
• For any element in the set, and for every real number, the product of the element and that real number is in the set.
• (these two conditions are called "closed under linear combinations," because the linear combinations of u and v are au+bv )

Note 1.1: Although on the real numbers these conditions are absurd, try looking at vectors (it is a "vector space"!). Consider <1,1,1> and <3,1,1> and notice that a vector space can be constructed around them so that <3,0,1> is not in that space. Also notice that it is not possible to construct a vector space so that <2,1,1> is not in that space.
Note 1.2: Notice that under these conditions, matrices are "vectors," because the set of all of them follow those rules.


Exercise 1:
Write the conditions of a vector space formally (that is, with the notations from Notation 0)

Prove that <2,1,1> is in any vector space with <1,1,1> and <3,1,1> (that is, find a and b so that <a,a,a> + <3b,b,b> = <2,1,1>). I know this is pretty easy but the point is to turn that first gibberish into the parenthetical statement - which is readable.

What element must all vector spaces contain?


Notation 1:
In higher mathematics, vectors in the traditional sense are usually written as columns: $\left({\begin{array}{c}1\\2\\-7\\4.1\end{array}}\right)$

The set of all vectors with n real components is called $\mathbb {R} ^{n}$. Such objects are n-dimensional. So that column vector up there... I'll just let that sink in.


Vocab 2:
A basis of a vector space $V$ is any smallest set $B$ of vectors (not numbers!) so that every element in the vector space is a linear combination of the elements of B. Read that very slowly until it makes sense. Then read this:
Formally $[\forall (v\in V)\rightarrow \exists (a_{1},a_{2},a_{3}...a_{n}\in \mathbb {R} )]\rightarrow a_{1}B_{1}+a_{2}B_{2}+...a_{n}B_{n}=v$. Notice that I assumed $B_{i}\in B$, and the size of B was n.

There are many sets that will satisfy B, but there is a collection of smallest sets. A set with n+1 elements over the same vector space is too large, and it implies that one of the vectors in the set is a linear combination of the others. If a set of vectors has this property, it is called linearly dependent, and if not it is called independent, (formally linearly independent, but independent is good enough). Sets of linearly dependent vectors are by definition not bases. However, if they satisfy the other property, B is said to span V.

The dimension of a vector space $V$ is the size of $B$. If you try and draw a vector space with dimension two you will find this has an intuitive meaning.


Vocab 3:
A bijection is a one-to-one function; sometimes this is said as "one to one and onto". Function has a messier definition that I will leave to Onto... but an intuitive definition of function will get you by - with one exception. Functions are no longer things that turn real numbers into real numbers, but now objects into objects. This is explained more in Notation 2.

A transformation or operator is a function. A linear operator $T$ is a function satisfying this identity (property): $T(au+bv)=aT(u)+bT(v)$, where a, b, u and v represent what they have so far.

I think you know about Function composition, but in case you forgot.


Notation 2:
$f:S_{1}\rightarrow S_{2}$ is read "f maps S1 to S2," or sometimes the more wordy "f sends elements of S1 to elements of S2." f is a function here. The important thing here is that it turns elements in $S_{1}$ into elements in $S_{2}$. The functions you are used to would be denoted $f:\mathbb {R} \rightarrow \mathbb {R}$. However, other things are not so foreign, you simply may not have thought of them as functions. For example the combination $_{n}C_{k}$ is a function. It looks like $C:\mathbb {N} ^{2}\rightarrow \mathbb {N}$. So it is a bit of an abstraction, but not horrible. Also notice that addition on, say, 15 dimensional vectors, is a function, $+:\mathbb {R} ^{30}\rightarrow \mathbb {R} ^{15}$. Do you see why? (It's not very practical to think about addition this way, but for the sake of argument)

$A\subset S$ is read "A is a subset of S." So A is also a set. Notice how this differs from $a\in S$.


Exercise 2:
Create a function, a rule, that sends natural numbers to high dimensional vectors (at least 5). Notate it.

Can vector spaces have maps from the reals to the reals (functions in the old sense) as elements? If so, find a vector space of such maps. If not, explain why. (Go back to the rules. What is a vector space?)

Let's define a "subspace" of $V$ as a set $W\subset V$ that is also a vector space. Explain why $\mathbb {R} ^{2}$ is not a subspace of $\mathbb {R} ^{3}$. Find a function $f:\mathbb {R} ^{2}\rightarrow \mathbb {R} ^{3}$ so that the set given by $f(u)$ is.