Brouwer Fixed Point Theorem and Matrix Eigenvalues

Proposition

A square matrix with all positive entries has a positive eigenvalue.

Proof.

Consider the standard simplex:

$$\begin{array}{}\text{(1)}& {\mathrm{\Delta }}^{n-1}:=\left\{\sum _{i=1}^n{\theta }_i{e}_i\right|{\theta }_i\ge 0,\text{and}\sum _{i=1}^n{\theta }_i=1\},\end{array}$$

which consists of points in ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ with ${\ell }^1$$\ell^1$-norm equal to 1.

For a matrix $A:{\mathbb{R}}^n\to {\mathbb{R}}^n$$A: \mathbb{R}^n \to \mathbb{R}^n$, define the function:

$$\begin{array}{}\text{(2)}& f(x)=\frac{Ax}{\parallel Ax{\parallel }_1},\end{array}$$

which maps points in ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ back to ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ (since the entries of $A$$A$ are all positive) and is a continuous function.

By the Brouwer Fixed Point Theorem:

Every continuous function from a nonempty convex compact subset $K$$K$ of a Euclidean space to $K$$K$ itself has a fixed point.

We know that $f(x)$$f(x)$ has at least one fixed point, i.e., there exists $x\in {\mathrm{\Delta }}^{n-1}$$x \in \Delta^{n-1}$ such that:

$$\begin{array}{}\text{(3)}& Ax=\parallel Ax{\parallel }_{1}x.\end{array}$$

From a Child’s Insight to Categorical Reflection: Subobjects and Quotient Objects in Group Theory and Abelian Categories

Background Story and a QuestionThe answer of the kidsAn answer from the relation between quotient object and subobject in category of GroupThe relation between quotient object and subobject in Abelian Category

This blog aims to show that an interesting fact, in some category, like $\mathrm{Grp}.$$\mathrm{Grp}.$

We have a canonical embedding: $(\mathrm{Quo}(\mathcal{X}),\le )\hookrightarrow (\mathrm{Sub}(\mathcal{X}),\subseteq )$$(\mathrm{Quo}(\mathcal X),\le)\hookrightarrow (\mathrm{Sub}(\mathcal X),\subseteq)$ for all $\mathcal{X}\in \mathrm{Ob}(\mathcal{C})$$\mathcal X\in\mathrm{Ob}(\mathcal C)$​.

That is, the poset of quotient object is a subobject of the poset of subobject in $\mathrm{Pos}$$\mathrm{Pos}$.

In Abelian Category, $(\mathrm{Quo}(\mathcal{X}),\le )\cong (\mathrm{Sub}(\mathcal{X}),\subseteq )$$(\mathrm{Quo}(\mathcal X),\le)\cong(\mathrm{Sub}(\mathcal X),\subseteq)$. Here $\mathrm{Sub}$$\mathrm{Sub}$ refer to subobject and $\mathrm{Quo}$$\mathrm{Quo}$ refer to quotient object, which is the dual concept of subobject. i.e., quotient object in $\mathcal{C}$$\mathcal C$ means subobject in ${\mathcal{C}}^{\mathrm{op}}$$\mathcal C^{\mathrm{op}}$​.

Background Story and a Question

Again, the initial idea of this blog comes from a quite stupid question. Let me explain the background.

Recently, Tsing Hua University is offering some public math course for kids in junior high school, one of the course is Topological Galois Theory. Here is the link of the course: Click here. So for junior high school kids, to understand Topological Galois Theory, they should know some basic fact about group theory, for example, what is simple group and what is the odd permutation/even permutation and commutator subgroup $[G,G]$$[G,G]$.

After he introduce what is commutator subgroup, Jianfeng Lin ask a question for kids: In permutation group $S_n$$S_n$​, could you find a element not in the commutator subgroup?

The answer of the kids

A kids whose voice hadn't even changed yet—he wasn't even in his voice-breaking period, think about 3 seconds and say: "odd permutation!" What a smart kids he is!

The idea is, the generator of commutator subgroup is $ab{a}^{-1}{b}^{-1}$$aba^{-1}b^{-1}$​, which is even permutation. So it is a junior high school level answer, I would like to give a different answer, a university level answer lol.

An answer from the relation between quotient object and subobject in category of Group

Let $G$$G$ be a group, then $\mathrm{Quo}(G)$$\mathrm{Quo}(G)$ is one-one corresponds to its normal subgroup.

This correspondence is order preserving map since:

$$\begin{array}{}\text{(1)}& K\subset N⟹\pi :G/K\to G/N\end{array}$$

The canonical projection comes from ${\displaystyle \frac{G/K}{N/K}}\cong G/N$$\dfrac{G/K}{N/K}\cong G/N$.

In the poset of quotient object, we could denote $\pi :G/K\to G/N$$\pi:G/K\to G/N$ as $G/K\le G/N$$G/K\le G/N$.

Hence we have $(\mathrm{Quo}(\mathcal{X}),\le )\hookrightarrow (\mathrm{Sub}(\mathcal{X}),\subseteq )$$(\mathrm{Quo}(\mathcal X),\le)\hookrightarrow (\mathrm{Sub}(\mathcal X),\subseteq)$.

Now we could answer the question:

In permutation group $S_n$$S_n$, could you find a element not in the commutator subgroup?

The answer is definitely yes, since $[G,G]$$[G,G]$ corresponding to the abelianlization of $G$$G$, and $G/[G,G]$$G/[G,G]$ is the smallest abelian quotient object of $G$$G$. i.e. for all $\pi :G\to G/H$$\pi:G\to G/H$ where $G/H$$G/H$ is abelian, there exists a unique factorization:

$$\begin{array}{}\text{(2)}& G\stackrel{{\pi }_1}{\to }G/[G,G]\stackrel{{\pi }_2}{\to }G/H\end{array}$$

Notice that the quotient object $\mathrm{sign}:S_n\to \mathbb{Z}/2\mathbb{Z}$$\mathrm{sign}:S_n\to\mathbb Z/2\mathbb Z$ is an abelian quotient object of $S_n$$S_n$​, hence we have

$$\begin{array}{}\text{(3)}& [G,G]\subseteq A_n\end{array}$$

Therefore, odd permutation is not in $[G,G]$$[G,G]$​​​​​.

Remark. You may ask me how do you now that $[G,G]$$[G,G]$ is a normal subgroup, indeed, we have a strong conclusion,

$[G,G]$$[G,G]$​ is a characteristic subgroup.

Definition.

A subgroup $H\subseteq G$$H\subseteq G$ is called characteristic subgroup if it is fixed under group automorphism, equivalently, restrict to $H$$H$ is an endomorphism, equivalently, group automorphism maps the generator of $H$$H$ to $H$$H$.

Proposition. Characteristic subgroup is normal subgroup.

Proof. Notice that $a⟼ga{g}^{-1}$$a\longmapsto gag^{-1}$ is a group automorphism as well. $◻$$\square$

Proposition. $[G,G]$$[G,G]$ is a characteristic subgroup.

Proof. We only need to that group automorphism $\phi$$\varphi$ maps the generators of $[G,G]$$[G,G]$ to $[G,G]$$[G,G]$.

It follows that $\phi [a,b]=[\phi (a),\phi (b)]$$\varphi[a,b]=[\varphi(a),\varphi(b)]$ directly. You do not expand $[a,b]$$[a,b]$ since it is purely $\times$$\times$ and inverse, preserved under group homomorphism. Just like $f(p(x))=p(f(x))$$f(p(x))=p(f(x))$ for a ring homomorphism $f$$f$. $◻$$\square$

This is not ture in $\mathrm{Ring}$$\mathrm{Ring}$ since ideal is not ring and $\mathrm{e}\mathrm{p}\mathrm{i}$$\rm epi$ is not means surjection. $\mathbb{Z}\to \mathbb{Q}$$\mathbb Z\to \mathbb Q$ is epi as well.

The relation between quotient object and subobject in Abelian Category

Let $\mathcal{A}$$\mathcal A$ be an abelian category,

1. Given a kernel $k$$k$, if we take its cokernel $c$$c$, then $k$$k$ is again the kernel of $c$$c$

2. Given a cokernel $c$$c$, if we take its kernel $k$$k$, then $c$$c$ is again the cokernel of $k$$k$.

3. Every monomorphism is a kernel and every epimorphism is a cockernel.

This three propositions give us a bijection between $\mathrm{Sub}(X)$$\mathrm{Sub}(X)$ and $\mathrm{Quo}(X)$$\mathrm{Quo}(X)$​.

Since we have that $c:\mathrm{Sub}(X)\to \mathrm{Quo}(X),k:\mathrm{Quo}(X)\to \mathrm{Sub}(X)$$c:\mathrm{Sub}(X)\to\mathrm{Quo}(X),k:\mathrm{Quo}(X)\to \mathrm{Sub}(X)$ such that $c\circ k={\mathrm{id}}_S,k\circ c={\mathrm{id}}_Q$$c\circ k=\mathrm{id}_{S},k\circ c=\mathrm{id}_Q$.

Also, $c,k$$c,k$ are order preserving, hence we have $(\mathrm{Sub}(X),\le )\cong (\mathrm{Quo}(\mathrm{X}),\subseteq )$$(\mathrm{Sub}(X),\le)\cong(\mathrm{Quo(X),\subseteq})$.

For example, in $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$ ...