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this is the last question I was not able to answer to prepare for my final.

Consider the linear transformation $T: \Bbb R^3 \to \Bbb R^2$ defined, for $x = (x_1, x_2, x_3) ∈ \Bbb R^3$ , by

$$T(x) = (x_2-x_3 \,,\,x_1+2x_2)$$

where $\Bbb R^3$ and $\Bbb R^2$ are endowed with the normal Euclidean norms.

Show that $T$ is bounded by finding a positive real number $M$ such that

$$||T(x)||≤ M||x|| \;\;\text{for all}\;\; x ∈ \Bbb R^3$$

Justify your answer.

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    $\begingroup$ You should not copy questions from your text book without at least showing some effort in where you ran into trouble, or without showing what you tried. $\endgroup$ – sxd Mar 18 '13 at 22:05
  • $\begingroup$ If you're preparing for a final I bet you can tell us what you tried, where did you get stuck, etc....? $\endgroup$ – DonAntonio Mar 18 '13 at 22:08
  • $\begingroup$ I am pretty sure $M=2013$ would work. $\endgroup$ – Julien Mar 18 '13 at 22:22
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    $\begingroup$ @DonAntonio, I just inspected all of his questions, seems like all of them are like these... $\endgroup$ – sxd Mar 18 '13 at 22:40
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Sami's solution is pretty nice. I present an alternative solution (of course in no way better or more elegant, but for a novice it might look a bit more convenient).

Please note, that it suffices to find an upper bound for $\|T(x)\|$ for all $x$ with $\|x\|=1$, because for arbitrary $x$ we have

$$ \begin{align} \|T(x)\| \leq M \|x\| \quad\Longleftrightarrow\quad \|T\left(\frac{x}{\|x\|}\right)\| \leq M \end{align} $$

where $\frac{x}{\|x\|}$ has norm 1. Now assume $\|x\|=1$, then each coordinate's absolute value is bounded by 1 and so

$$ \begin{align} \|T(x)\|^2 &= (x_2-x_3)^2 + (x_1 + 2x_2)^2 \\ &= 5x_2^2 + x_3^2 + x_1^2 - 2x_2 x_3 + 4x_2x_1 \\ &\leq 5 + 1 + 1 + 2 + 4 = 13 \end{align} $$

It follows that $\sqrt{13}$ satisfies the desired property.

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  • $\begingroup$ I assumed that $\mathbb{R}^3$ and $\mathbb{R}^2$ are endowed with the Euclidean norm. At least, that's what the OP wrote - or what are you referring to? $\endgroup$ – thomas Mar 18 '13 at 22:52
  • $\begingroup$ You're right, of course. Thanks. +1 $\endgroup$ – DonAntonio Mar 18 '13 at 22:53
  • $\begingroup$ Should it not be 5+1+1 - 2+4=9? $\endgroup$ – Omar Mar 19 '13 at 19:28
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    $\begingroup$ No, because we have $|x_2|<1$ and $|x_3|<1$, hence $|x_2x_3|<1$ and especially $-x_2x_3<1$, but NOT $-x_2x_3<-1$ $\endgroup$ – thomas Mar 19 '13 at 22:34
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The norm $1$ is defined by $$||(x,y,z)||_1=|x|+|y|+|z|.$$ and by Cauchy-Shwarz inequality we have $$||(x,y,z)||_1\leq\sqrt{3}||(x,y,z)||.$$ Now we have $$||T(x)||^2=(x_2-x_3)^2+(x_1+2x_2)^2\leq ||x||_1^2+||2x||_1^2=5||x||_1^2\leq 15||x||^2$$ hence $$||T(x)||\leq \sqrt{15}||x||.$$

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    $\begingroup$ Why using the $1$ norm? Also, it should be $\leq \sqrt{3}$ by CS. $\endgroup$ – Julien Mar 18 '13 at 22:27
  • $\begingroup$ @julien I used the $1$ norm just for the sake, in fact, I've only used the notation that it seems lighter. $\endgroup$ – user63181 Mar 19 '13 at 5:50

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