<s>[INST] If $T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$. Determine $||T||$. Define $T : \mathbb{R}^2 → \mathbb{R}$ ( $\mathbb{R}^2$ & $\mathbb{R}$ being
equipped with the Euclidean norm) by
$T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$.
Determine $||T||$.

My thoughts:- 
We know that
$ \qquad \left\|{T}\right\| = \sup \left\{{\left|{Th}\right|: \left\|{h}\right\| = 1}\right\}$
here $h=(x,y)$ and $||h||=\sqrt{x^2+y^2}=1$
So we need to find out the maximum value of $2x+y$ with the condition that $x^2+y^2=1$.
Am I right?
 [/INST] Yes, you are right. To do so, note that 
$$ T'(x,y) = \begin{pmatrix} 2 & 1 \end{pmatrix} $$
and for $F(x,y) = x^2 + y^2 - 1$ we have 
$$ F'(x,y) = \begin{pmatrix} 2x & 2y \end{pmatrix} $$
We hence have to solve
\begin{align*}
  2 &= 2x \lambda\\
  1 &= 2y\lambda\\
  x^2 + y^2 &= 1
\end{align*} 
So $x = 1/\lambda$, $y = 1/2\lambda$, hence $y = \frac x2$. Plugging this into the third equation gives
$$ x^2 + \frac{x^2}4 = 1 \iff x^2 = \frac 45 \iff x = \pm \frac 2{\sqrt 5} $$
This gives $y = \pm \frac 1{\sqrt 5}$. Now evaluate $T$ at the two points.
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