Is Linear transformations $T_1,T_2 : \mathbb{R}^n\rightarrow \mathbb{R}^n$ Invertible? Let $T_1$ and $T_2$ be two Linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$.
Let $\{x_1,x_2,\cdots,x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1(x_i)\neq 0$ for every $1\leq i\leq n$ and that $x_i\perp Ker (T_2)$ for every $1\leq i\leq n$. 
Which of the following is true?


*

*$T_1$ is invertible

*$T_2$ is invertible

*Both $T_1$ and $T_2$ are invertible

*Neither $T_1$ nor $T_2$ is invertible.


As $T_1(x_i)\neq 0$ for each $1\leq i\leq n$ we do not have $T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 $ unless each $a_i=0$ i.e.,$a_1x_1+a_2x_2+\dots+a_nx_n=0$ i.e., $T_1$ is one one thus invertible.
I am not sure if $T_2$ is invertible or not.
we have $x_i\perp Ker (T_2)$ for all $x_i$. Would that be a good idea say something like
$\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)$ and as they span whole space we would have 
$\mathbb{R}^n\perp Ker(T_2)$ and thus $Ker(T_2)=0$ so it is injective so it is invertible...
Thus both $T_1$ and $T_2$ are invertible?
I would be so thankful if someone can assure what it has been done here is sufficient/clear.
THank you

You are wrong that $T_1$ is invertible; it need not be. Indeed for $n=2$, $\{x_1,\ldots,x_n\}$ the standard basis, and $T_1$ given by the matrix $$\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$$ neither of $T_1(x_1)=x_1-x_2$ nor $T_1(x_2)=x_2-x_1$ is zero, but $T_1(x_1+x_2)=0$. Moreover one can take $T_2=I$ to satisfy $x_i\perp\ker(T_2)=\{0\}$ for $i=1,2$.
Your argument that $\ker(T_2)=\{0\}$ always, hence $T_2$ is invertible, is correct.
