2

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?

  1. $T_1$ is invertible.
  2. $ T_2$ is invertible.
  3. Both $T_1, T_2$ are invertible.
  4. Neither $ T_1$ nor $ T_2$ is invertible.
Chappers
  • 67,606
SKarantha
  • 437

1 Answers1

1

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.

aexl
  • 2,102