persistent homology group as vector subspace A persistent homology group is defined as $i^\ast (H_k(X^i))$ where $i$ is the function $i^\ast:H_k(X^i)\to H_k(X^j)$ for any $i<j$.
All my homology groups have coefficient in a field $K$ so they all are vectorial space. How can I say that the persistent homology group is also a vector space(Edited)?
I know it's a subset of a vector space, but I don't know how to show that it's a vector space.
 A: I would refer you to Sections 3.1 and 3.2 of the paper "Computing Persistent Homology" by Afra Zomorodian and Gunnar Carlsson, available at https://link.springer.com/article/10.1007/s00454-004-1146-y or https://geometry.stanford.edu/papers/zc-cph-05/zc-cph-05.pdf. Indeed, you have a persistence module $H_k(X^1)\to H_k(X^2)\to \ldots \to H_k(X^{n-1})\to H_k(X^n)$ where all homology groups are taken with coefficients in a field $K$. Therefore, as explained in Sections 3.1 and 3.2, this persistence module is isomorphic to a (graded) module over $K[t]$, where $t$ acts by "mapping forward one step." Note that the field $K$ is a subring of $K[t]$, which means that this persistence module is also a module over $K$, i.e. a vector space over the field $K$.
What is the vector space you get? This vector space is obtained simply by ignoring the extra structure given by the action of $t$. When you ignore this action by $t$, you have lost all of the "persistent" information. The resulting vector space is just the direct sum $\bigoplus_{i=1}^n H_k(X^i)$, which is a vector space over the field $K$ since each summand $H_k(X^i)$ is a vector space over the field $K$. From the perspective of persistence, this vector space isn't so interesting. To recover the persistence information, you instead want to consider this persistence module as a (graded) module over $K[t]$, since the action by $t$ (again see Sections 3.1 and 3.2) explains how the different homology groups $H_k(X^i)$ map to each other as $i$ increases.
