prove that specific linear transformation is bounded 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.
 A: 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.
A: 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 
$$||T(x)||\leq \sqrt{15}||x||.$$
