Are linear transformations $T_1$, $T_2$ invertible? Let $T_1, T_2$ be two linear transformations from $\mathbb{R}^n$ to  $\mathbb{R}^n$. Let $\{ x_1, x_2,....x_n\}$ be a basis of  $\mathbb{R}^n$. Suppose that $T_1  x_i \neq 0$ for every $i= 1,2,...,n$ and that $x_i \perp $Ker $ T_2 $ for every $i= 1,2,...,n.$ Which of the following is/ are necessarily true?


*

*$T_1$ is invertible.  

*$ T_2$ is invertible. 

*Both $T_1, T_2$ are invertible. 

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

 A: We define the linear transformation $T_1: \Bbb R^n \to \Bbb R^n$ as 
$$\begin{align*}
T_1(x_1) &:= x_1 - x_2 \; ,\\
T_1(x_2) &:= x_2 - x_3 \; ,\\
& \; \; \; \vdots \\
T_1(x_{n-1}) &:= x_{n-1} - x_n \; ,\\
T_1(x_n) &:= x_{n} - x_1 \; .
\end{align*}$$
Then we have the condition $T(x_i) \neq 0$ fullfilled for $i = 1, \ldots, n$.
Let's define $T_2: \Bbb R^n \to \Bbb R^n$ as the identity map $T_2(x_i) = x_i$ for $i = 1, \ldots, n$. Then $\operatorname{ker}(T_2) = \{ 0 \}$, so $x_i \perp \operatorname{ker}(T_2)$ is also fullfilled for each $i = 1, \ldots, n$. Now observe that 
$$ T_1\left( \sum_{i=1}^n x_i \right) = \left( \sum_{i=1}^{n-1} x_i - x_{i+1} \right) + x_n - x_1 = 0 \; ,$$
i.e. $\sum_{i=1}^n x_i \in \operatorname{ker} (T_1)$, so $T_1$ is not invertible. Thus the point 1. is not necessarily true.
Let $T_1, T_2: \Bbb R^n \to \Bbb R^n$ be two linear transformations, such that the given conditions are fullfilled. Assume that $T_2$ is not invertible. Then $\operatorname{ker}(T_2) \neq \{ 0 \}$. Let $0 \neq v \in \operatorname{ker}(T_2)$. It follows, that $\{x_1, \ldots, x_n, v\}$ is linearly independant, but this is a contradiction to $\dim(\Bbb R^n) = n$. So $T_2$ always needs to be invertible, and the point 2. is necessarily true.
