---
# System prepended metadata

title: TDA_TEAM_5

---

---
title: TDA_TEAM_5
---

Team:
Seth
Larry

**Homologies: $H_i=ker(\partial_i)/im(\partial_{i+1})$**


$$\dim H_i=\dim~ker(\partial_i)-\dim~im(\partial_{i+1})$$ 

**Example A.** 
1. Create the chain.
$$0\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12\}$$

3. Create matrices to describe boundary maps.
$$\partial_1:\quad \begin{bmatrix}1&1&0\\1&0&1\\ 0&1&1\\0&0&0\end{bmatrix}\sim \begin{bmatrix}1&0&1\\0&1&1\\ 0&0&0\\0&0&0\end{bmatrix}$$

4. Compute homologies. 
$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})=4-2=2$$ 

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})=1-0=1$$ 

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

**delta 2**
The image of $\partial_{2}$ is 0 because it's a linear transformation from the $\mathbf{0}$ linear space

**delta 0**

The image of $\partial_{0}$ is $0$
The $ker(\partial_{0})$ = $C_{0}$
The dimension of the kernel is dim $ker(\partial_{0})) = 4$

**delta 1**

The dimension of the kernel is dim $ker(\partial_{1})) = 1$
The dimention of the image is dim $im(\partial_{1})) = 2$

 3 $\in$ $H_{0}$
 
 There are two clusters, since dim $H_{0}$ = 2. The clusters consist of the sets of vertices {3} and {0,1,2}

**Example B.** 
1. Create the chain.
$$0\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0
$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12,23\}$$

3. Create matrices to describe boundary maps.
$$\partial_1:\quad \begin{bmatrix}1&1&0&0\\1&0&1&0\\ 0&1&1&1\\0&0&0&1\end{bmatrix}\sim \begin{bmatrix} 1& 1& 0& 0\\ 0&1 &1 & 0\\ 0 &0 &0 & 1\\ 0&0 &0 &0 \end{bmatrix}
$$

4. Compute homologies. 
$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})= 4 - 3 = 1
$$ 

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})= 1 - 0 = 1$$ 

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

**delta 2**
The image of $\partial_{2}$ is 0 because it's a linear transformation from the $\mathbf{0}$ linear space

**delta 0**

The image of $\partial_{0}$ is $0$
The $ker(\partial_{0})$ = $C_{0}$
The dimension of the kernel is dim $ker(\partial_{0})) = 4$

**delta 1**

The dimension of the kernel is dim $ker(\partial_{1})) = 1$
The dimention of the image is dim $im(\partial_{1})) = 3$

 
 There is one cluster, since dim $H_{0}$ = 1. The cluster consist of the set of vertices {0,1,2,3}

**Example C.** 
1. Create the chain.
$$0\quad \rightarrow\quad C_2\quad \rightarrow\quad C_1\quad \rightarrow\quad C_0\quad \rightarrow\quad 0$$

2. Describe the simplicial complexes

$$C_0:=\{0,1,2,3\},\quad C_1:=\{01,02,12,13,23\},\quad C_2:=\{012\}$$

3. Create matrices to describe boundary maps.
$$\partial_1:\quad \begin{bmatrix} 1 & 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 1 & 0\\ 0&1&1&0&1\\ 0&0&0&1&1\end{bmatrix}\sim \begin{bmatrix} 1&1&0&0&0\\0&1&1&1&0 \\0&0&0&1&1  \\0&0&0&0&0 \end{bmatrix}$$

$$
\partial_{2}: \quad \begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1 \\
\end{bmatrix}
$$

4. Compute homologies. 

$$\dim H_0=\dim~ker(\partial_0)-\dim~im(\partial_{1})= 4 - 3 = 1
$$ 

$$\dim H_1=\dim~ker(\partial_1)-\dim~im(\partial_{2})= 2 - 1 = 1$$

$$\dim H_2=\dim~ker(\partial_2)-\dim~im(\partial_{3})= 1 - 0 = 1$$

5. Describe your findings in your own words. AS MUCH AS YOU CAN.

The dimension of $H_{0}$ suggests there is 1 cluster.

The dimension of $H_{1}$ suggests there is 1 loop.

The dimension of $H_{2}$ suggests there is 1 void.



