$T:V \to V$ if $B = \{v_1,...,v_n\}$ is a basis for $V$ what about $\{Tv_1,...Tv_2\}$ I see a prove which uses something like this: 
Let $T:V \to V$ be a linear transformation. 
Let $B = \{v_1,...,v_n\}$ be a base for $V$.
Can i conclude that $\{Tv_1,..,Tv_n\}$ is also a base for $V$. 
If so, what about the case when $kerT \neq \{0\}$? 
And how do we prove it? 
Thanks. 
 A: if $\{v_{1} , v_{2},..., v_{n}\}$ is a basis of $\ V$ and  $ \ T:\ V \rightarrow V$ a linear transformation then obviously  $\{T(v_{1}) , T(v_{2}),..., T(v_{n})\}$ is a generating set for $Im(T)$ ie: $Im(T)=vect(T(v_{1}) , T(v_{2}),..., T(v_{n}))$ now we know that:
$$\dim(Ker(T))+dim(Im(T))=dim(V)$$
thus having $Ker(T)=\{ 0_{\ V} \}\implies dim(Im(T))=dim(V)\implies V=Im(T)$ 
A: *

*Suppose that $\ker(T)=\{0\}$.
Let $a_1,\ldots,a_n$ be such that $\sum_{i=1}^n a_i T(v_i)=0$ and set $w=\sum_{i=1}^n a_i v_i$. By linearity of $T$, we have
$$0= \sum_{i=1}^n a_i T(v_i)= \sum_{i=1}^n T(a_i v_i)=T(w).$$
As $\ker(T)=\{0\}$, $T(w)=0$ implies that $w=0$ and thus $a_1=\ldots=a_n=0$. Therefore $T(v_1),\ldots,T(v_n)$ is linearly independent. Finally, $n=\dim(V)$ implies that $T(v_1),\ldots,T(v_n)$ is a basis of $V$.

*Suppose that $\ker(T)\neq \{0\}$. Let $u\in\ker(T), u \neq 0$. Let $b_1,\ldots,b_n$ be such that $u=\sum_{i=1}^n b_i v_i$. Then, we have 
$$0=T(u) = \sum_{i=1}^n T(b_i v_i)=\sum_{i=1}^n b_i T(v_i).$$
As $u\neq 0$ not all $b_i$ are $0$ and thus $T(v_1),\ldots,T(v_n)$ is not linearly independent. Hence, $T(v_1),\ldots,T(v_n)$ is not a basis of $V$.
A: All of the answers assume that $V$ is finite dimensional. Injectivity does not guarantee that the image of a basis is still a basis when $V$ is infinite dimensional. 
For example, let $V= \ell^2$, $v_i = (0,...,1,0,...)$ be the standard Schauder basis, and $T$ be the right shift operator defined by $(v_1, v_2,...) \mapsto(0,v_1,v_2,...)$. This is clearly injective and linear but $(1,0,...)$ is not in the image of $T$.
