Practical question on computation of Betti numbers Maybe my question has an answer here, but I don't understand that. I will pose mine differently. The following definitions are from [1, Sec 4.1].
Consider a simplicial complex. The $k$-th chain group $C_k$ is the set of all $k$-chains $$\sum_i n_i \, \text{[oriented $k$-simplex $i$]}$$ with integer coefficients $n_i$. The boundary homomorphism $\partial_k : C_k \to C_{k-1}$ between chain groups is defined by 
$$
\partial_k [v_0, \ldots, v_k] = \sum_i (-1)^i [v_0, \ldots, \hat{v}_i, \ldots, v_n].
$$
The $k$-th homology group is $$H_k := \ker \partial_k / \operatorname{im} \partial_{k+1}.$$
As a finitely generated group, it is isomorphic to a direct sum of primary cyclic groups and infinite cyclic groups, or it decomposes into into the torsion subgroup and the free subgroup. The Betti number $\beta_k$ is the rank of the free subgroup.
Let $A_k \in \mathbb{R}^{m \times n}$ be the matrix representation of $\delta_k$ with respect to some basis. Consider it as an operator between the vector spaces $\mathbb{R}^n$ and $\mathbb{R}^m$ (over the reals). We can compute $$\tilde{\beta}_k := \dim \ker A_k - \dim \operatorname{im} A_{k+1}.$$
I think this is the dimension of the homology group obtained when in the chain group we admit with real coefficients(?).
How are $\beta_k$ and $\tilde{\beta}_k$ related?
[1] Zomorodian, Compuational topology, Notes, 2009.
 A: Better notation would be $A_k$ for what you just call $A$.
Then indeed $\dim\ker A_k-\dim\text{Im}\,A_{k+1}$ is $\beta_k$.
This follows from the Universal Coefficient Theorem for homology,
but can probably be proved from first principles.
ADDED IN EDIT
I outline a more naive approach.
I'm assuming that the simplicial complex is finite. The matrix $A_k$
represents a map $\partial_k:C_k\to C_{k-1}$ as well as a linear
map $\overline\partial_k:C_k\otimes\Bbb R\to C_{k-1}\otimes\Bbb R$.
The groups $C_k$ are free Abelian. Therefore the kernel and image of
$\partial_k$ are free Abelian groups. Each free Abelian group has
a rank and so we may consider $\text{rk}\ker(\partial_k)$ and
$\text{rk}\,\text{im}(\partial_k)$. Now these are the same as the
nullity and rank of the linear map $\overline\partial_k$. This
follows since the rank of an integer matrix is the same as the rank
of the $\Bbb Z$-span of its rows (to be pretentious one might say that
$\Bbb R$ is a flat $\Bbb Z$ modules). From this observation,
$\beta_k=\tilde\beta_k$ is immediate.
A: I'll try showing $\beta_k=\tilde\beta_k$ by expanding out the relevant portions of the proof of the universal coefficient theorem.
The first thing is let's define $A_k$ instead to be the $\mathbb{Z}$-valued matrix for $\partial_k:C_k\to C_{k-1}$, with respect to the basis of oriented simplices.  Assume the simplicial complex has only finitely many simplices for this matrix to be finite-dimensional.  Let $\tilde{A}_k$ be the same matrix, but thought of as being $\mathbb{R}$-valued.  So, using $\operatorname{rank}$ to mean free rank,
\begin{align}
\beta_k&=\operatorname{rank}(\ker A_k/\operatorname{im} A_{k+1})\\
\tilde{\beta}_k&=\dim(\ker \tilde{A}_k/\operatorname{im} \tilde{A}_{k+1})=\dim\ker\tilde{A}_k-\dim\operatorname{im}\tilde{A}_{k+1}
\end{align}
(where the latter equality is just the rank-nullity theorem).
A main way to change coefficients in algebra is tensor products.  $\mathbb{Z}^n\otimes\mathbb{R}\cong\mathbb{R}^n$.  If you are not familiar with the concept, just take this axiomatically for now.  One can say that $\tilde{A}_k=A_k\otimes \mathbb{R}$ to represent the change of the coefficient ring.  Tensor product properties:


*

*$\mathbb{Z}\otimes\mathbb{R}\cong\mathbb{R}$.

*$(\mathbb{Z}/n\mathbb{Z})\otimes\mathbb{R}\cong 0$.

*If $A,B$ are abelian groups, $(A\oplus B)\otimes\mathbb{R}\cong (A\otimes\mathbb{R})\oplus(B\otimes\mathbb{R})$.

*So: If $A$ is a finitely generated abelian group, $A\otimes\mathbb{R}\cong \mathbb{R}^{\operatorname{rank} A}$.


One part to this is that we have a chain complex with $\mathbb{Z}$-coefficients
$$\cdots\xrightarrow{A_{k+1}} \mathbb{Z}^{n_k}\xrightarrow{A_k} \mathbb{Z}^{n_{k-1}}\xrightarrow{A_{k-1}}\cdots$$
which can be converted to $\mathbb{R}$-coefficients by tensoring with $\mathbb{R}$ to get
$$\cdots\xrightarrow{\tilde A_{k+1}} \mathbb{R}^{n_k}\xrightarrow{\tilde A_k} \mathbb{R}^{n_{k-1}}\xrightarrow{\tilde A_{k-1}}\cdots$$
Now, the difficulty is to find a way to relate the homology groups for each of these chain complexes.  Let $H_k$ denote the $k$th homology group of the $C_k$ complex.  By definition of $H_k$, there are short exact sequences
$$0\to \operatorname{im} A_{k+1}\hookrightarrow \ker A_k\to H_k\to 0$$
Since $\operatorname{im} A_{k+1}$ and $\ker A_k$ are free abelian groups, $(\operatorname{im} A_{k+1})\otimes\mathbb{R}\cong \operatorname{im}\tilde A_{k+1}$ and $(\ker A_k)\otimes\mathbb{R}\cong \ker \tilde A_k$, and the inclusion map tensored with $\mathbb{R}$ remains injective.  This ends up meaning there is a short exact sequence
$$0 \to \operatorname{im}\tilde A_{k+1} \hookrightarrow \ker \tilde A_k \to H_k\otimes \mathbb{R}\to 0$$
By the first isomorphism theorem, $H_k\otimes \mathbb{R}\cong \ker \tilde A_k / \operatorname{im}\tilde A_{k+1}$.  This latter quotient is the definition of the homology with real coefficients, so $$\tilde\beta_k=\dim(H_k\otimes\mathbb{R})=\dim(\mathbb{R}^{\operatorname{rank} H_k})=\operatorname{rank} H_k=\beta_k$$
