To compute the simplicial homology group for $\mathbb{S}^{1}$, Hatcher uses below triangulation:
From this triangulation, we have one vertex $v$ and one edge $e$. Thus, we have $C_{0}(X)=\mathbb{Z}$ and $C_{1}(X)=\mathbb{Z}$, which gives us the simplicial chain complex $$C_{1}(X)=\mathbb{Z}\longrightarrow_{\partial_{1}}C_{0}(X)=\mathbb{Z}.$$
The only thing to do is to compute $\partial_{1}$. To do so, let us write $e=\langle V_{0}, V_{1}\rangle$, with the understanding in mind that $V_{0}=V_{1}=v$, and thus we have $$\partial_{1}e=\partial_{1}\langle V_{0}, V_{1}\rangle=\langle V_{0}\rangle-\langle V_{1}\rangle=v-v=0.$$
This implies that $Im(\partial_{1})=0$ and $\ker(\partial_{1})=C_{1}(X)=\mathbb{Z}.$
Hence, $$H_{0}(X)=\ker(\partial_{0})/{Im(\partial_{1})}={C_{0}(X)}/{0}=\mathbb{Z}$$ and $$H_{1}(X)={\ker(\partial_{1})}/{Im(\partial_{2})}={\mathbb{Z}}/{0}=\mathbb{Z}.$$
However, I am trying to prove that the homology group is the same if we triangulate $\mathbb{S}^{1}$ in different ways.
The triangulation of $\mathbb{S}^{1}$ is just to divide the circle into different numbers of $1-$simplex. For instance, see the graph below: we use $5$ vertices to divide this circle into five sections of $1-$simplexes.
Thus, more generally, if use $n$ vertices to divide the circle, we have $n$ sections of $1-$simplexes. Thus, $C_{0}(X)=\mathbb{Z}^{n}$ and $C_{1}(X)=\mathbb{Z}^{n}$.
But then I got confused. If we used the same computation as above, we would see that $$\partial_{1}(any \ edges)=0.$$ Therefore, $$Im(\partial_{1})=0,\ \text{and}\ \ \ker(\partial_{1})=C_{1}(X)=\mathbb{Z}^{n}.$$
Then, $H_{0}(X)=\mathbb{Z}^{n}$ and $H_{1}(X)=\mathbb{Z}^{n}$.
What is happening here? Do we still have one vertex and one edge, even though I triangulate the circle with $n$ vertices? why?
Thank you!
Edit 1: (Mistakes above)
As comments pointed out, $\partial_{1}(any\ edges)\neq 0$, so we need to compute it as follows:
Now, let us order the vertices and edges as follows: pick the lowest vertex as $v_{0}$, then order them counterclockwise as $v_{0}, v_{1},\cdots, v_{n-1}$. Then the edges (arcs) are $$e_{1}=\langle v_{0}, v_{1}\rangle,\ \ e_{2}=\langle v_{1},v_{2}\rangle,\ \ \cdots,\ \ e_{n-1}=\langle v_{n-2}, v_{n-1}\rangle,\ \ \text{and}\ \ e_{n}=\langle v_{n-1}, v_{0}\rangle.$$
This implies that $$C_{0}(X)=\langle v_{0}, v_{1},\cdots, v_{n-1}\rangle=\mathbb{Z}^{n}$$ and $$C_{1}(X)=\langle e_{1}, e_{2},\cdots, e_{n}\rangle=\mathbb{Z}^{n}.$$
Then for each $1\leq k<n$, we have $$\partial_{1}e_{k}=\partial_{1}\langle v_{k-1},v_{k}\rangle=v_{k-1}-v_{k},$$ and for $k=n$ we have $$\partial_{1}e_{n}=\partial_{1}\langle v_{n-1}, v_{0}\rangle=v_{n-1}-v_{0}.$$
Therefore, the image of $\partial_{1}$ is spanned by the sets $\{v_{0}-v_{1}, v_{1}-v_{2},\cdots, v_{n-2}-v_{n-1}, v_{n-1}-v_{0} \}$, whose matrix representation is an $n\times n$ matrix
$$
\begin{pmatrix}
1 &-1&0&0&\cdots&0&0&0&0\\
0 &1&-1&0&\cdots&0&0&0&0\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
0&0&0&0&\cdots&0&0&1&-1\\
-1&0&0&0&\cdots&0&0&0&1.
\end{pmatrix}
$$
Note that $v_{0}-v_{1}+v_{1}-v_{2}+\cdots+v_{n-2}-v_{n-1}+v_{n-1}-v_{0}=0$ and thus the above generating set is not linearly independent, and $v_{n-1}-v_{0}=v_{0}-v_{1}+v_{1}-v_{2}+\cdots+v_{n-2}-v_{n-1}$, so the basis of $Im(\partial_{1})$ is actually $\{v_{0}-v_{1},v_{1}-v_{2},\cdots, v_{n-2}-v_{n-1}\}$ and thus $$Im(\partial_{1})=\langle v_{0}-v_{1},v_{1}-v_{2},\cdots, v_{n-2}-v_{n-1}\rangle=\mathbb{Z}^{n-1}.$$ Indeed, they are linearly independent since $$a_{1}(v_{0}-v_{1})+a_{2}(v_{1}-v_{2})+\cdots+a_{n-1}(v_{n-2}-v_{n-1})=0$$ if and only if $$a_{1}v_{0}+(a_{2}-a_{1})v_{1}+\cdots+(a_{n-2}-a_{n-1})v_{n-1}=0$$ if and only if $$a_{1}=a_{2}=\cdots=a_{n-1}=0.$$
Therefore,
\begin{align*}
H_{0}(X)={ker(C_{0}(X)\longrightarrow 0)}/{Im(\partial_{1})}&={C_{0}(X)}/{Im(\partial_{1})}\\
&={\langle v_{0}, v_{1},\cdots, v_{n-1}\rangle}/{\langle v_{0}-v_{1},v_{1}-v_{2},\cdots, v_{n-2}-v_{n-1}\rangle}
\end{align*}
What is this quotient? I am not sure if it is generally true that $\mathbb{Z}^{n}/\mathbb{Z}^{n-1}=\mathbb{Z}$, so is there any way to rewrite the basis for $Im(\partial_{1})$ so that the quotient is obvious?
Also,
\begin{align*}
\partial_{1}(a_{1}e_{1}+a_{2}e_{2}+\cdots+a_{n}e_{n})&=a_{1}(v_{0}-v_{1})+a_{2}(v_{1}-v_{2})+\cdots+a_{n-1}(v_{n-2}-v_{n-1})+a_{n}(v_{n-1}-v_{0})\\
&=(a_{1}-a_{n})v_{0}+(a_{2}-a_{1})v_{1}+\cdots+(a_{n-1}-a_{n-2})v_{n-2}+(a_{n}-a_{n-1})v_{n-1}.\\
&=0
\end{align*}
if and only if $a_{1}=a_{2}=\cdots=a_{n}$, so what is the kernel???
Best Answer
I don't understand why you're bothering with the matrix. Kernel and image can be computed pretty easily without it (at least in this case).
Let $c = \sum a_i e_i$ be a 1-chain in $\ker \partial_1$, then $0 = \partial_1 \left(\sum a_i e_i \right) = \sum a_i (v_{i-1} - v_{i}) = \sum (a_{i+1} - a_i)v_i$.
(I am being imprecise with the indices here: write $a_{n+1} = a_{0}$ etc.)
By freeness of $C_0$, we get that for each $i$, $a_i = a_{i+1}$. So, $a_1 = a_2 = \dots = a_n$.
Hence, $\ker \partial_1 \cong \mathbb{Z}$ with generator $\sum e_i$. So, $H_1(S^1) \cong \Bbb Z$.
Now, as you noted, $\text{im } \partial_1 = \text{span}\{ v_0 - v_1, \dots, v_{n-1} - v_0\}$. This set is not linearly independent. Consider the set $\{ v_1 - v_0, v_2 - v_0, \dots, v_{n-1} - v_0\}$. It is not difficult to show that this set spans $\text{im }\partial_1$ and is linearly independent. So, $\text{im} \partial_1 = \Bbb Z^{n-1}$. Hence, $H_0(S^1) = \text{coker }\partial_1 \cong \Bbb Z^n/\Bbb Z^{n-1} \cong \Bbb Z$.
But, if you insist on using matrix, note that if you replace each row by the sum of all the rows above it(including that row itself), then you will obtain the row-echelon form of the matrix, in which the last row is zero. Hence, nullity = 1 and rank = n-1.