"There is hardly any theory which is more elementary [than linear algebra], in spite of the fact that generations of professors and textbook have obscured its simplicity by preposterous calculations with matrices." - Jean Dieudonné

Consider the following problem: 

We have two vector spaces $U, V$; a linear transformation $T: U \rightarrow V$ and we want to represent it by a matrix. Since this process has an idea that can be easily grasped, it is common to omit the details, although they can be somewhat confusing if you try to do it in full extension. We proceed in doing the full process formally:

Definition: Given two vector spaces $U, V$, of finite dimensions $n$ and $m$ respectively; a linear transformation $T: U \rightarrow V$; a basis $\{e_j\}$ of U and a basis $\{f_i\}$ of V, we define the matrix associated to T, the domain basis $\{e_j\}$ and the codomain $\{f_i\}$ as the matrix with coefficients $a_{ij}$, where $a_{ij}$ are the numbers such that:

$$\displaystyle T(e_j)=\sum_{i=1}^m a_{ij}f_i$$

(note that everything is well-defined, since we are talking about bases)

We denote this matrix by ${}_{f_i}M^{T}_{e_j}$.

Definition: Let $\{e_j\}$ and $\{\bar{e_j}\}$ be bases of $U$. The change of basis B from $\{e_j\}$ to $\{\bar{e_j}\}$  is the linear transformation that takes $\{e_j\}$ to  $\{\bar{e_j}\}$ . The matrix ${}_{e_j}M^{B}_{e_j}$ is called change of basis matrix.

Theorem 1: ${}_{f_i}M^{T}_{\bar{e_j}}={}_{f_i}M^{T}_{e_j}.{}_{e_j}M^{B}_{e_j}$.

Proof: $\displaystyle T(e_j)=\sum_{i=1}^m a_{ij}f_i$
Let $\displaystyle \bar{e_j}=\sum_{i=1}^n b_{ij} e_i$.

Then: $\displaystyle T(\bar{e_j})=T(\sum_{i=1}^n b_{ij} e_i)=\sum_{i=1}^n b_{ij} T(e_i)=
\sum_{i=1}^n b_{ij} \sum_{k=1}^m a_{ki}f_k=\sum_{k=1}^m (\sum_{i=1}^n a_{ki} b_{ij})f_k=\sum_{i=1}^m (\sum_{k=1}^n a_{ik} b_{kj})f_i$

Since $\displaystyle \sum_{k=1}^n a_{ik} b_{kj}$ is the coefficient $c_{ij}$ of the product of matrices, and the matrix $b_{ij}$ is clearly the change of basis matrix from $\{e_j\}$ to $\{\bar{e_j}\}$, the result follows. $\blacksquare$

The following lemma and corollary will be useful:

Lemma: Given $T_1: U \rightarrow V$, $T_2: V \rightarrow W$, let $T:=T_2 \circ T_1$. Then:
$${}_{g_i}M^{T}_{e_k}={}_{g_i}M^{T_2}_{f_j}.{}_{f_j}M^{T_1}_{e_k}$$

Proof: Similar computation as above. $\blacksquare$


Corollary: ${}_{e_j}M^{T}_{e_j}=({}_{e_j}M^{T^{-1}}_{e_j})^{-1}$


Proof: Take $T_1=T$ and $T_2=T^{-1}$ in the lemma above. $\blacksquare$


Now, we come to the codomain problem of changing basis:


Theorem 2: ${}_{\bar{f_i}}M^{T}_{e_j}={}_{\bar{f_i}}M^{B^{-1}}_{\bar{f_i}}
{}_{\bar{f_i}}M^{T}_{e_j}$

Proof: $\displaystyle T(e_j)=\sum_{i=1}^m a_{ij}f_i$

Let $\displaystyle f_i=\sum_{k=1}^n b_{ki} \bar{f_k}$.

Then:  $\displaystyle T(e_j)=\sum_{i=1}^m a_{ij}\sum_{k=1}^n b_{ki} \bar{f_k}=\sum_{i=1}^m \sum_{k=1}^n a_{ij} b_{ki} \bar{f_k}=
\sum_{k=1}^m \sum_{i=1}^n b_{ki} a_{ij} \bar{f_k}=\sum_{i=1}^m \sum_{k=1}^n b_{ik} a_{kj} \bar{f_i}$

Since $\displaystyle \sum_{k=1}^n b_{ik} a_{kj}$ is the coefficient $c_{ij}$ of the product of matrices, and the matrix $b_{ij}$ is clearly the change of basis matrix from $\{\bar{f_i}\}$ to $\{f_i\}$, the result follows. $\blacksquare$

Corollary: ${}_{\bar{f_i}}M^{T}_{e_j}=({}_{\bar{f_i}}M^{B}_{\bar{f_i}})^{-1}
{}_{\bar{f_i}}M^{T}_{e_j}$

Proof: Follows from the previous theorem and previous corollary. $\blacksquare$

Now, with Theorems 1 and 2, we arrive at:

Theorem (Pre-Change of Basis): Given a linear transformation from a vector space $U$ of dimension $n$ to itself, we have:

$${}_{\bar{e_j}}M^{T}_{\bar{e_j}}=({}_{\bar{e_j}}M^{B}_{\bar{e_j}})^{-1}.
{}_{e_j}M^{T}_{e_j}.{}_{e_j}M^{B}_{e_j}$$

Why "Pre-Change of Basis"? Note there is a pesky annoyance on the right side that shouldn't be there: $({}_{\bar{e_j}}M^{B}_{\bar{e_j}})^{-1}$ should be $({}_{e_j}M^{B}_{e_j})^{-1}$. How to fix this? Well, we don't fix. We prove it is fixed:


Lemma: Given $T:U \rightarrow U$ isomorphism,

$${}_{e_j}M^{T}_{e_j}={}_{T(e_j)}M^{T}_{T(e_j)}$$


where $T(e_j)$ should be understood as the basis of $U$ given by $\{T(e_j)\}_{j=1}^n$


Proof: $\displaystyle T(e_j)=\sum_{i=1}^m a_{ij}e_j$


and


$\displaystyle T(T(e_j))=\sum_{i=1}^m a_{ij}T(e_j)$ $\blacksquare$.

Now, we have the theorem:

Theorem: Given a linear transformation from a vector space $U$ of dimension $n$ to itself, we have:

$${}_{\bar{e_j}}M^{T}_{\bar{e_j}}=({}_{e_j}M^{B}_{e_j})^{-1}.
{}_{e_j}M^{T}_{e_j}.{}_{e_j}M^{B}_{e_j}$$

and, passing to its common form:

Theorem (Change of Basis): Given a linear transformation from a vector space $U$ of dimension $n$ to itself, we have:

$${}_{e_j}M^{T}_{e_j}={}_{e_j}M^{B}_{e_j}.{}_{\bar{e_j}}M^{T}_{\bar{e_j}}.({}_{e_j}M^{B}_{e_j})^{-1}$$

Note that in the case of a matrix with orthonormal basis of eigenvectors, the notation for this equality is commonly seen as:

$$A=P.D.P^{-1}$$

So, what is a quotient topology? Well, the idea is that we will "identify points". More colourfully, we will shrink a whole subset (or whole subsets) to a point (or points). For example: when you take the disk in $\mathbb{R}^2$ and identify all the points of the boundary (the 1-dimensional sphere $S^1$), you get the 2-dimensional sphere $S^2$:










When you take a square, and identify a point of each side with its "opposite", you get a torus.

Ok, so now we go to the theory. (First, let me remark the following: I will assume that the equivalence between a "partition of a set" and "equivalence classes" of some equivalence relation is well-understood. To every partition of a set, we can define the equivalence class: "$x \sim x'$ if $x$ is in the same subset of $x'$". And every equivalence class determines a partition of a set.)

With that in mind, let's begin:

Definition: If $X$ is a topological space and we have a partition of $X$, call $X^*$ the set of those subsets that form the partition. Give $X^*$ the following topology: A subset of $X^*$ (which is a collection of subsets of X) is open if and only if the union of that collection is open in $X$. This is the quotient space.

Equivalently, if $X$ is a topological space and we have an equivalence relation $\sim$, call $X/ \sim$ the quotient space (meaning, the set of equivalence classes). We then have the map: $\pi :X \rightarrow X / \sim$ that takes $x$ to its equivalence class $\bar{x}$. Now, give  $X/ \sim$ the following topology: $U'$ is open in $X/ \sim$ if and only if $\pi^{-1}(U')$ is open in $X$. This is the quotient space.


It may not be easy to see that they are equivalent at a first glance if you are not familiar with equivalence relations, or with topology, but a bit of thought will make it clear. I will adopt the latter definition in the calculations (but will write $X/ \sim $ as $X^*$ for convenience) since it is more algebraic, hence, easier to handle.

Now, before introducing the next useful theorem, let me give some acquaintance to commutative diagrams. Consider the isomorphism theorem:

Isomorphism Theorem: Given an homomorphism $f: G \rightarrow H$ between two groups, we have $\displaystyle G/\ker(f) \simeq Im(\phi)$.

What this theorem says can be given more precision, in the following language:

Isomorphism Theorem: Every homomorphism $f: G \rightarrow H$ between two groups induces an injective homomorphism $\bar{f}: G / \ker(\phi) \rightarrow H$  making the following diagram commute (meaning: $f= \bar{f} o \pi$) 











Take a minute to understand this and then we can proceed:


Proposition: Given $X, Y$ topological spaces, a continuous $f:X\rightarrow Y$ which is constant on every equivalence class induces a continuous map $f^*: X^* \rightarrow Y$ making the following diagram commute:

Demonstration: Well, recall the proof of the isomorphism theorem: what happens is that, when you define the map, you must show that it independs of the representative of the class for it to be well-defined. In that case, that happens because the quotient is on the kernel. In this case, it is even simpler: $f$ is constant in the equivalence class!
Define $f^*(\bar{x})=f(x)$. It is well-defined by the previous observation, and obviously $f=f^*o\pi$. It remains to show that $f^*$ is continuous. Lets show pre-image of open is open! But, a set $U$ is open in $X^*$ if and only if $\pi^{-1} U$ is open in $X$. So, we have to show that, for every open $V$ in $Y$, $\pi^{-1} o f^{*-1} (V)$ is open. But this set is exactly $f^{-1}(V)$. Since $f$ is continuous, we have proved $f^*$ is continuous.

Let's see an application of this and prove... that the circle is a segment where you identify the endpoints!







Consider $I:=[0,2\pi]$, and $S^1:=(\cos(t),\sin(t))$, for $t \in [0,2\pi)$. Now, make the following equivalence relation in $I$: every point is equivalent only to itself, except $0$ and $2\pi$, which are equivalent (they are in the same equivalence class). Therefore, we have $I^*$

Take the map $f: I \rightarrow S^1$ that takes $x$ to $(\cos(x), \sin(x))$. Note that $f(0)=f(1)$. So, $f$ satisfies the hypothesis of our preceding proposition (namely, it is constant in every class!). Then, we have an induced continuous map $f^*: I^* \rightarrow S^1$. (Note that $f^*$ is bijective). Now, take the following map: $g: S^1 \rightarrow I^* $ that takes $(\cos(x), \sin(x))$ to $\bar{x}$). It is obviously continuous on all points, except possibly at $f(1,0)$ (since all other classes are points). But the neighbourhoods of $f^*(1,0)=\bar{1}=\bar{0}$, now that we are at the quotient space, must "contain" a neighbourhood of $1$ and a neighbourhood of $0$. Why? Suppose you have a neighbourhood $\bar{U}$ of $\bar{1}$. So, $\pi^{-1}(\bar{U})$ must be an open set containing $0$ and $1$. Therefore, there are two small intervals around $0$ and $1$ in $\pi^{-1}(\bar{U})$, which will be taken to the corresponding classes. So, this $g$ is continuous, since we can take a small enough piece of the circle around $(1,0)$ for which the function will fall inside those intervals. But this $g$ is the inverse of $f^*$. So, we found a homeomorphism. We could also do something more elegant (but would need more knowledge about topology): Since $I$ is compact, so is $I^*$. But $f^*$ is bijective, and $S^1$ hausdorff, so $f^*$ is in fact a homeomorphism.

In Pt.3, we shall talk about another style of glueing things.

When I was studying topology, I kinda disregarded the Quotient Topology part. It was short, but I didn't see much use of it inside the theory we were looking at during the course. Turns out I was right, but the conclusion - "so, disregard it" turned out to be quite wrong, since the knowledge is both aesthetic and useful.

Well, to begin with, I think it's a good idea to explain what is a Topological Space.  Well, consider you are in a metric space. The open sets of the space are sets $A$ such that each point of $A$ has a ball around it still inside of $A$. We will say any open set around a point is a neighbourhood of the point. Open sets have two interesting properties:

1 - Arbitrary union of open sets is still open.
2- Finite intersection of open sets is still open.

Turns out these two properties are enough to demonstrate a good amount of results. Just like when we leave $\mathbb{R}^n$ and go to metric spaces, having realized that (for a lot of purposes) all we need is the triangle inequality (plus some good conditions) to talk meaningfully of distance, we will abandon metric spaces, desiring to talk meaningfully of neighbourhoods (open sets). This generalization is, in my point of view, a good motivation by itself. But the fact is that topology has far-reaching uses in a lot of fields of mathematics, and helps us understand and formalize things like "a circle is a closed interval with its endpoints glued".

So, what is a Topological Space?

Definition: A Topological Space is a set $X$, together with a set $\tau$ of subsets of $X$, such that:

i) $\emptyset \in \tau$; $X \in \tau$
ii) Given a family $A_{\alpha}$ in $\tau$,   $\displaystyle \bigcup A_{\alpha} \in \tau$
iii) $A_1, A_2 \in \tau \Rightarrow A_1 \cap A_2 \in \tau$

The elements of $\tau$ are called open sets of the topology.

Examples:

i) Metric Space - A metric space is a topological space, where the open sets are precisely its open sets (note that the first "open sets" refers to "the open sets of the topology", whereas the second "open sets" refers to the "open set as defined in a metric space).
ii) Given a set $X$, the topology $\tau=\{\emptyset, X\}$ is called the trivial topology. It is easily verified that it is a topology.
iii) Given a set $X$, the topology $\tau=\mathcal{P} (X)$ is called discrete topology. It is also easily verified that it is a topology.

For an example that is new to first-readers of topology:

iv) Given a set $X$, consider $\tau=\{\text{subsets of } X \text{ that have finite complement, and the empty set}\} $. This is a topology, called the finite complement topology, and a simple-to-describe one. Moreover, it has interesting features. First, let's define a property that is kind of reasonable to expect:

Definition: We say the topological space $(X,\tau)$ is Hausdorff  if, for every two points $p_1, p_2$ of $X$, there exists a neighbourhood $U_1$ of $p_1$ and a neighbourhood $U_2$ of $p_2$ such that $U_1 \cap U_2=\emptyset$. 

Essentially, this says we can "properly" separate points. It is obvious a metric space is a Hausdorff topological space. In fact, every topological space should be Hausdorff... or not?

Take our example (iv): Consider the set $\mathbb{Z}$ with the finite complement topology. Take the points $1$ and $2$. A neighbourhood $U_1$ of $1$ is, for example, $\{1,2,6,7,8,9,10,...\}$ and a neighbourhood $U_2$ of $2$ is, for example, $\{2,3,4,1210,1430,1431,1432,1433,1434,...\}$. But it doesn't matter what neighbourhoods of $1$ and $2$ you use, since they always have finite complement, there is a number $N_1$ such that any $n>N_1$ will be in $U_1$. Likewise, there is a number $N_2$ such that any $n>N_2$ will be in $U_2$. So, the intersection will always be non-empty ($\max\{N_1, N_2\}+1$ will be in the intersection, for example). Therefore, $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

Well, now that we know what is a topological space, let's introduce one of the most fundamental concepts of Topology: Continuous Functions. As we will see, continuous functions with a continuous inverse will behave quite analogous to what a "isomorphism" is in linear algebra.

So, let's begin with the analysis definition of continuity:

Definition 1 of continuity: $f:X\rightarrow Y$ is said to be continuous at $x_0$ if $\forall \epsilon > 0 \exists \delta > 0 : d(x,x_0)<\delta \Rightarrow d(f(x),f(x_0))<\epsilon$.

There is, in this definition, an explicit use of distance. Here comes one that is easy to see to be equivalent, but quite more appealing:

Definition 2 of continuity: $f:X\rightarrow Y$ is said to be continuous at $x_0$ if for all neighbourhood $V$ of $f(x_0)$, there exists a neighbourhood $U$ of $x_0$ that is taken inside $V$ by $f$.

In fact, this definition is automatically generalized to topological spaces, where there is no distance. Now, we have the following proposition:

Proposition: A function $f:X \rightarrow Y$ is continuous if and only if, for every open set $V$ of $Y$, $f^{-1}(V)$ is open. 



Note that this can be rephrased more compactly with: Inverse image of open is open.


Demonstration: Let's prove $\Rightarrow$:
Take an open set $V$ of $Y$. We have to prove $f^{-1}(V)$ is open. So, take a point $p$ in $f^{-1}(V)$. By continuity, there is a neighbourhood $U$ of $p$ that goes into $V$. So, this implies $U \subset f^{-1}(V)$. Therefore, $p$ is an interior point. Since $p$ is arbitrary, $f^{-1}(V)$ is open.
For $\Leftarrow$, 
let $p$ be a point of X. We will prove $f$ is continuous at $p$. Let $V$ be a neighbourhood of $f(p)$. By hypothesis, $f^{-1}(V)$ is open. Hence, there is a neighbourhood $U$ of $p$ in $f^{-1}(V)$. So, this neighbourhood will be taken to $V$. Therefore, $f$ is continuous at $p$. $\square$

Turns out that, for topological purposes, this property of continuity is so important that it is common to define  "continuous" as "pre-image of open is open". In fact, if a continuous function is bijective and has a continuous inverse (a homeomorphism), it is easy to see that it will take open sets from one topology to open sets of another topology. Hence, homeomorphisms "preserve the topology" in some sense. If there is a homeomorphism between two topological spaces, we can say they are topologically equivalent.

This "Pt.1" serves as an introduction to "Pt.2", where I'll talk about quotient spaces, and adjunction spaces.