Norm of a linear operator from $\mathbb{R}^2\to\mathbb{R}^3$

Question

Find $\|T\|_{\mathcal{L}}$ where $T\in\mathcal{L}(\mathbb{R}^2,\mathbb{R}^3)$ is defined by $$T(\mathbf{x})=(x,2x, 3x)\ \ \ \forall \mathbf{x}=(x,y)\in\mathbb{R}^2$$ My approach

1. $T$ is a linear operator (easy to prove);
2. $\|\mathbf{x}\|=\sqrt{x^2+y^2}\ \ \ \forall \mathbf{x}\in\mathbb{R}^2$
3. $\|T(\mathbf{x})\|=\sqrt{x^2+4x^2+9x^2}=\sqrt{14}|x|$

so

$\frac{\|T(\mathbb{x})\|}{\|\mathbf{x}\|}=\frac{\sqrt{14}|x|}{\sqrt{x^2+y^2}}\le\frac{\sqrt{14}|x|}{|x|}=\sqrt{14}\ \ \ \forall\mathbf{x}\ne \mathbf{0}$

This means that $\|T\|_{\mathcal{L}}\le\sqrt{14}$. For $\mathbf{x}=(1,0)$ we have that $\|T(\mathbf{x})\|=\sqrt{14}$ so $\|T\|_{\mathcal{L}}=\sqrt{14}$.

Second approach

Let $\mathbf{x}$ be a unitary vector of $\mathbb{R}^2$, such that $$\|\mathbb{x}\|=1\implies x^2+y^2=1 \ \ \ \to |x|=\sqrt{1-y^2}\ \ \ \forall y\in [-1,1]$$

so

$$\|T(\mathbf{x})\|=\sqrt{14}|x|=\sqrt{14}\sqrt{1-y^2}\le\sqrt{14}\ \ \ \forall y\in [-1,1]$$

In particular $\|T(\mathbf{x})\|=\sqrt{14}$ for $(x,y)=(1,0)$, hence $\|T\|_{\mathcal{L}}=\sqrt{14}$.

Are these solutions ok? Thanks.

Answer 1

Yes, your solutions are correct (with the caveat $x\neq 0$ vs. $\textbf{x}\neq 0$ given in the comments).

A more conceptual approach is the following: Your linear operator is given by the matrix $A:=\begin{pmatrix}1&0\\2&0\\3&0\end{pmatrix}$. The operator norm induced by the euclidean norm is given by the square root of the spectral radius of $A^T A=\begin{pmatrix}14&0\\0&0\end{pmatrix}$.