Vtyjkyuk

Find $|T|_mathcalL$ where $TinmathcalL(mathbbR^2,mathbbR^3)$ is defined by $$T(mathbfx)=(x,2x, 3x) forall mathbfx=(x,y)inmathbbR^2$$
My approach

1. $T$ is a linear operator (easy to prove);


2. $|mathbfx|=sqrtx^2+y^2 forall mathbfxinmathbbR^2$


3. $|T(mathbfx)|=sqrtx^2+4x^2+9x^2=sqrt14|x|$

so

$fracmathbfx=fracxsqrtx^2+y^2lefracx=sqrt14 forallmathbfxne mathbf0$

This means that $|T|_mathcalLlesqrt14$. For $mathbfx=(1,0)$ we have that $|T(mathbfx)|=sqrt14$ so $|T|_mathcalL=sqrt14$.

Second approach

Let $mathbfx$ be a unitary vector of $mathbbR^2$, such that
$$|mathbbx|=1implies x^2+y^2=1 to |x|=sqrt1-y^2 forall yin [-1,1]$$

so

$$|T(mathbfx)|=sqrt14|x|=sqrt14sqrt1-y^2lesqrt14 forall yin [-1,1]$$

In particular $|T(mathbfx)|=sqrt14$ for $(x,y)=(1,0)$, hence $|T|_mathcalL=sqrt14$.

Are these solutions ok? Thanks.

Yes, your solutions are correct (with the caveat $xneq 0$ vs. $textbfxneq 0$ given in the comments).

A more conceptual approach is the following: Your linear operator is given by the matrix $A:=beginpmatrix1&0\2&0\3&0endpmatrix$. The operator norm induced by the euclidean norm is given by the square root of the spectral radius of $A^T A=beginpmatrix14&0\0&0endpmatrix$.