Exercise 11 page 30 in Functional Analysis book of Conway The following is Exercise 11 page 30 in Functional Analysis book of Conway:

If $A = 
        \begin{bmatrix}
        a & b \\
        c & d  \\
        \end{bmatrix}
$, put $\alpha=[|a|^2+|b|^2+|c|^2+|d|^2]^{1/2}$ and show that $\|A\| = \frac12 (\alpha^2+\sqrt{\alpha^4-4\delta^2})$, where ${\delta}^2 = \det A^* A$.

Here some details are added but there is no satisfactory answer. Also here is another link that asks the same question without an answer based on materials taught. The exercise appears at  page 30 of the book which lacks concepts like :

The operator Euclidean norm of $A$ is just the square root of the largest eigenvalue of $A^∗A$.

A detailed simple answer or a hint would be much appreciated.
 A: First, there is a typo in this exercise: the result to be proved is $\|A\|^2 = \frac12 (\alpha^2+\sqrt{\alpha^4-4\delta^2})$. Just check with  $A = 
        \begin{bmatrix}
        2 & 0 \\
        0 & 2  \\
        \end{bmatrix}
$.
Now, it is easy to see that
$$\|A\|^2=\sup{\{\|Ax\|^2, \|x\|=1}\} = \sup{\{\langle Ax, Ax \rangle, \|x\|=1}\} = \\ =\sup{\{\langle A^*Ax, x \rangle, \|x\|=1}\}  \tag{1}$$
Since $A^*A$ is a self-adjoint operator, it is diagonalisable, with real eigenvalues. So
$$ \sup{\{\langle A^*Ax, x \rangle, \|x\|=1}\} = \text{ the largest eigenvalue of } A^*A \tag{2}$$
Now, let us compute $A^*A$. We have
$$ A^*A =  \begin{bmatrix}
        \bar a & \bar c \\
        \bar b & \bar d  \\
        \end{bmatrix}  
        \begin{bmatrix}
        a & b \\
        c & d  \\
        \end{bmatrix}= 
       \begin{bmatrix}
        |a|^2 +|c|^2 &  \bar a b + \bar c d \\ 
        a \bar b + c \bar d & |b|^2 +|d|^2  \\
        \end{bmatrix}
$$
So, the eigenvalues of $A^*A$ will be the roots of the the equation
$$ \lambda ^2 - (|a|^2 +|c|^2  + |b|^2 +|d|^2 ) \lambda + \det(A^*A)= 0 $$
Now since we defined $\alpha^2= |a|^2 +|c|^2  + |b|^2 +|d|^2$ and $\delta^2 = \det(A^*A)$, we have that
$$ \lambda ^2 - \alpha^2 \lambda + \delta^2= 0 $$
So,
$\lambda  = \frac12 (\alpha^2\pm\sqrt{\alpha^4-4\delta^2})$. Since $A^*A$ has real eignevalues we know that $\alpha^4-4\delta^2 \geq 0$ and that the largest eigenvalue is $ \frac12 (\alpha^2 +\sqrt{\alpha^4-4\delta^2})$. So, from $(1)$ and $(2)$, we have that
$$\|A\|^2 = \frac12 (\alpha^2 +\sqrt{\alpha^4-4\delta^2})$$
