Norm of complex Matrix with euclidean norm Would you help me to prove the norm of 2 by 2 matrix 
\begin{equation}
A=\begin{bmatrix}
a & b \\ 
c & d%
\end{bmatrix}
\end{equation}
is $$\def\trc{\operatorname{tr}}\frac{\trc(A^*A)+\sqrt{\trc(A^*A)^2-4\det(A^*A)}}{2}.$$ I just try by maximizing $h(x,y)=||A\begin{bmatrix}
x \\ 
y%
\end{bmatrix}||$ where $x^2+y^2=1$ but not get this result.
 A: I think you are talking about the operator norm induced by the euclidean norm, and not the euclidean norm per se. In that case, the norm will be the square root of the number you mention. 
The operator norm of $A$ can be characterized as the square root of the biggest eigenvalue of $A^*A$. The eigenvalues of $A^*A$ are the roots of its characteristic polynomial. As $A^*A$ is $2\times 2$, its characteristic polynomial is $p(t)=t^2-\mbox{tr}(A^*A)\,t+\det(A^*A)$. So its biggest eigenvalue is 
$$
\frac{\mbox{tr}(A^*A)+\sqrt{\mbox{tr}(A^*A)^2-4\det(A^*A)}}2
$$
(the polynomial $p$ has always non-negative roots, so the formula above certainly gives the biggest eigenvalue). So the norm of $A$ is 
$$
\left(\frac{\mbox{tr}(A^*A)+\sqrt{\mbox{tr}(A^*A)^2-4\det(A^*A)}}2\right)^{1/2}.
$$
A: It's the spectral radius of $A^*A$, that is, the largest modulus of the eigenvalues of $A^*A$. We can express them with a second degree equation satisfied by the eigenvalues, and knowing they are non-negative, we can see that the given expression is the norm of the matrix.
