<s>[INST] Norm of a linear transformation Let $T:\mathbb R^2\to \mathbb R^2$ be given by the matrix $\begin{pmatrix}a&b\\ c& d\end{pmatrix}$.  Let $u:=a^2+b^2+c^2+d^2+2(ad-bc)$ and $v:=a^2+b^2+c^2+d^2-2(ad-bc)$. 
I need to show that $\mid\mid T\mid\mid=\frac 12 (\sqrt u+\sqrt v)$. 
I tried using the definition $\mid\mid T\mid\mid =\sup\{\mid \mid Tx\mid\mid\ : x\in \mathbb R^2, \mid\mid x\mid\mid=1 \}$, and trying to express each $x$ as $(\cos \theta,\sin \theta)$ for $\theta\in  [0,2\pi)$.  This just led to a big mess, though, eventually reaching that 
$$\mid\mid T(\cos\theta,\sin \theta)\mid\mid^2 < a^2+b^2+c^2+d^2+2(ab+cd)$$  
This is kind of similar to $u$ and $v$ but not quite good enough, so I'm not sure now if this approach with $\theta$ and the trig functions is the right one. 
I also don't know how to get started for the other inequality. 
 [/INST] You have to find largest eigenvalue of $T^tT$ and show that it is equal to $\frac{u+v+2\sqrt{uv}}{4}.$ You get into the following equations $$(x-a^2-c^2)(x-b^2-d^2)-(ab+cd)^2=0\\\implies x^2-(a^2+b^2+c^2+d^2)x+(ad-bc)^2=0\\\implies x_\max=\frac{a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2-4(ad-bc)^2}}{2}\\=\frac{u+v+2\sqrt{uv}}{4}.$$
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