Let $T:\mathbb{R}^2 \to \mathbb{R}^2$ be defined as $T\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{cc} 1 & 2\\ 2 & 1 \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right)$.
Prove that $\|v\|\le\|T(v)\|\le9\|v\|$, for every $v\ = \left(\begin{array}{c} x\\ y \end{array}\right)\in \mathbb{R}^2$. as $\|v\|$ is the norm of the vector $v$ in $\mathbb{R}^2$ defined by the standard inner product.
My Attempt:
as the transformation matrix $\left(\begin{array}{cc} 1 & 2\\ 2 & 1 \end{array}\right)$ is symmetric, we conclude that $T$ is Hermitian ($T=T^{*}$), which also means that $T$ is a normal transformation. I've tried to use quadratic forms as I can't understand exactly in what way one should handle the norm of a matrix.
the quadratic form I found for $T$ is:
$q(x_1,x_2)= x_1^2 +4x_1 x_2 + x_2^2$.
suppose $y_1 = x_1 + 2x_2, y_2 = x_2$ and $q(y) = y_{1}^2 -3y^2_2 $, hence by the change of bases we conclude that $q(x)$ and the transformation $T$ can be represented by the diagonal matrix:
$$A = \left(\begin{array}{cc} 1 & 0\\ 0 & -3 \end{array}\right).$$
How - or can we continue from here?