Show that the given linear transformation is invertble

Question

Let $T_1,T_2$ be two linear transformations from $\Bbb R^n$ to $\Bbb R^n$.Let $\{x_1,x_2\ldots x_n\}$ be a basis of $\Bbb R^n$ .Suppose that $T_1x_i\neq 0\forall i=1,2,.....n$ and that $x_i\perp \ker T_2\forall i=1,2,.....n$.

Which of the following are true?

1. $T_1$ is invertible
2. $T_2$ is invertible
3. $T_2,T_1$ is invertible
4. Neither $T_1$ nor $T_2$ is invertible

For $T_1$:

Define $T_1:\Bbb R^2\to \Bbb R^2$ by $T_1(x_1)=x_1;T_1(x_2)=-x_1\implies T_1(x_1+x_2)=0\implies T _1$ is not one-one and hence not invertible.

For $T_2$:

Let $x\in \ker T_2\implies Tx=0$ where $x=\sum_{i=1}^n c_ix_i$

Now $\langle x,x_i\rangle =0\forall i\implies c_i=0\forall i\implies x=0\implies \ker T_2=\{0\}$

**But I am unable to show that $T_2$ is surjective. Let $y\in \Bbb R^n\implies y=\sum c_ix_i$.But I don't know what steps to follow next.

Please give some hints.

Answer 1

Hint: what can you say about the range of the image of the basis by $T_2$?