Norm of a $2\times 2$ matrix as a Hilbert space operator Work in the Hilbert space $\mathbb C^2$. Let 
$$A = 
        \begin{bmatrix}
        a & b \\
        c & d  \\
        \end{bmatrix}
$$
be a matrix with entries in $\mathbb C$, and let $A$ act in the standard way on $\mathbb C^2$. This gives a linear operator on $\mathbb C^2$.
Define $\alpha=[|a|^2+|b|^2+|c|^2+|d|^2]^{1/2}$ and $\delta = \sqrt{\det A^* A}$. I would like to show that 
$$\|A\|=\frac{\alpha^2+\sqrt{\alpha^4-4\delta^2}}{2},$$
where $\|A\|$ is the norm as a Hilbert space operator. 
(Note the resemblance to the quadratic formula.)
This problem comes from page 30 of Conway's A Course in Functional Analysis, 2nd edition. It is problem 1.11 in chapter 2. 
 A: Hint: $\|A\|^2$ is the greatest eigenvalue of $A^* A$.
A: I will use without proof a few factors about operators on Hilbert spaces and their adjoints. Proofs can be found in the book referenced in the question. 
First, it is true that $\|A\|^2=\|A^*A\|$. This is useful because $A^*A$ is self-adjoint, so we can use a special characterization of the operator norm that applies only to self-adjoint operators. We have $\langle Ah, h\rangle \subset \mathbb R$ for all $h\in\mathbb C^2$, and 
$$\|A^*A\|=\sup\{ |\langle Ah, h \rangle|: \|h\|=1\}.$$
Recall that $A^*A$ is self-adjoint, so it is diagonalizable, and it has a basis of eigenvectors $v_1$ and $v_2$. Then any $h$ can be written and a linear combination of multiples of $v_1$ and $v_2$, and using the characterization of the norm above and the linearity of the inner product, the problem reduces to finding the largest eigenvalue of $A^*A$, as Robert Israel pointed out. 
You can then find the eigenvalue in the usual way by forming the characteristic polynomial, solving it using the quadratic equation, and taking the largest root. This explains why the formula given in the problem statement resembles the quadratic formula.
This same method also allows the derivation of formulas for the norms of $3\times 3$ and $4\times 4$ matrices, but not for $n\times n$ matrices with $n\ge 5$ (because there's no formula for the solution of a general $n$th degree equation with $n\ge 5$).
