Find norm of linear operator subordinate to Euclid norm

Question

Let $A$ be the linear operator in the space $V=\mathbb R^2$ with action $(x,y)\to(\frac{x}{2},\frac{-y}{3})$.
I have to find the norm of this operator subordinate to Euclid norm ($p=2$).

1. I found the matrix of operator.

2. I found square of the matrix of operator.

What should I do next?
By definion of norm, I find the sup of norm after operator action and divide by norm before?
Is the answer is $\frac{1}{6}$?
Sorry for terrible english.

Answer 1

The operator norm is defined as $\sup \{\|A(x,y)\|: \|(x,y)|| \leq 1\}$. This is then supremum of the numbers $\sqrt {\frac {x^{2}} 4+\frac {y^{2}} 9}$ over $(x,y)$ such that $x^{2} +y^{2} \leq 1$. The maximum value is attained when $x=1, y=0$ and the maximum value is $\frac 1 2$. So the norm is $\frac 1 2$.