Prove $T$ is a linear map on $V$, and the null space and range space are both finite, then vector space $V$ is finite.
Prove T is a linear map on V, and the null space and range space are both finite, then vector space V is finite.
My attempt:
The question asks to prove null $T$ and range $T$ are finite dimension $\implies$ V is finite dimension. I try to prove the inverse negation of this statement
i.e.V is infinite dimension $\implies$ null $T$ or range $T$ are infinite dimension
Since $V$ is an infinite-dimensional vector space, it has a list of basis vectors with infinite length. Then,based on the fundamental theorem of linear map,it's possible to find a basis of $V$ $(v_1,..,u_1,...)$ and a linear map $T$ such that $Tv_i=0$ (for $i\in \mathbb N$), and $Tu_j$ ($j\in \mathbb N$) are basis vectors for range $T$. This implies the dimension of null $T$ and range $T$ are both infinite.
Similarly, if the dimension of null $T$ is finite, it's possible to find an infinite list of vectors that, after applying the linear map, will become the basis for the range $T$ and if the dimension of range $T$ is finite, then it's also possible to find an infinitely long list of vectors as the basis vectors for null $T$.
However, it's impossible to have both null $T$ and range $T$ are infinite dimensions. Since two finitely long list vectors adding together can't become a infinitely long list of vectors that is the basis for vector space $V$
The idea of my proof is based on $V=$null $T \oplus$ $U$ and $U$ has a basis $(u_1,..,u_n)$ and $(Tu_1,...,Tu_n)$ is the basis for range $T$.
However, I'm suspicious of my proof. Thus, any suggestions?
Solution 1:
Let $e_1,...,e_m$ be a basis of $Ker(T)$ and $T(f_1),...,T(f_n)$ be a basis of $Range(T)$. Let $v$ be any element of $V$. Then $T(v)=\sum\alpha_i T(f_i)=T(\sum \alpha_i f_i)$. So $v-\sum \alpha_i f_i\in Ker T$. So $v-\sum \alpha_i f_i=\sum\beta_j e_j$. Hence $V$ is spanned by $e_1,...,e_m,f_1,...,f_n$ and $\dim(V)$ is finite. In fact one can prove that $e_1,...,e_m, f_1,...,f_n$ are linearly independent and $\dim V=m+n$.