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$T:\mathbb{R}^n\rightarrow \mathbb{R}$ 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.

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Well I guess you are right or you are wrong. (Include your work and thoughts up until now!) –  rschwieb Jun 13 '12 at 20:08
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Did you mean the range of $T$ to be $\mathbb{R}^n$? If not, is $T^2(x) = T(x)\cdot T(x)$? –  Jason DeVito Jun 13 '12 at 20:18
    
Is $T : \mathbb{R}^n \to \mathbb{R}^n$ or if not, what do you mean by $T^2$? –  dtldarek Jun 13 '12 at 20:22
    
$T(T-\lambda I)=0$ so $T(Tx-\lambda Ix)=0$ $||T(T-\lambda I)x||=0$ so $||T||=|\lambda| ||x||$ –  Une Femme Douce Jun 13 '12 at 20:22

2 Answers 2

up vote 1 down vote accepted

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.

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Hint:

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

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If the 2 was up there, then you would have $T^2=0$. –  copper.hat Jun 13 '12 at 20:35

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