2
$\begingroup$

$T:\mathbb{R}^n\rightarrow \mathbb{R}$ such that $T^2=\lambda T$ for some $\lambda\in\mathbb{R}$ Which of the following are true

  1. $||T(x)||=|\lambda| ||x||$ $\forall x\in\mathbb{R}^n$

  2. If $||Tx||=||x||$ for some nonzero vector $x$, then $\lambda=+1$ or $\lambda=-1$

  3. $T=\lambda I$

  4. $||Tx||>||x||$ for some non zero $x$ then $T$ must be singular.

Well, I guess 1 and 2.

  • 0
    $T(T-\lambda I)=0$ so $T(Tx-\lambda Ix)=0$ $||T(T-\lambda I)x||=0$ so $||T||=|\lambda| ||x||$2012-06-13

2 Answers 2

1

I don't think any of the statements in the problem hold true, since a counterexample can be found for each one. For (1), let

$T = \left( \begin{matrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right)$

Then $T^2 = 6T$ and $T:\mathbb{R}^n \rightarrow \mathbb{R}$. However,

$e_1 = \left( \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right), e_2 = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right) $

produces $T(e_1) = e_1, T(e_2) = 2e_2$. So (1) is not true. Similarly, (3) is not true since $T^2 \ne \lambda I$ for any $\lambda\in\mathbb{R}$. For statement (2), let

$ T = \left( \begin{matrix} 2 & 1 \\ 0 & 0 \\ \end{matrix} \right), e_2 = \left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right) . $

Then $T^2 = 2T$, but $||T(e_2)|| = ||e_2||$, disproving (2). Finally, you can disprove (4) by considering the matrix $T = \left(2\right)$, which is nonsingular. Hence, all statements are false.

2

Hint:

Consider $T = \left[\begin{matrix}2&0\\0&0\end{matrix}\right].$

  • 0
    If the 2 was up there, then you would have $T^2=0$.2012-06-13