I am reading this book (page 215) and I found in the bottom of this page an interesting relation that I want to use. The relation is
$$||A||_2 = \sqrt{\frac{||A||_F^2 + \sqrt{||A||_F^4 – 4 (\mathrm{det}\:A)^2}}{2}},$$ where
$$A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$ is a $2×2$ real matrix. The relation can be verified by using the definition of the spectral norm of $A$, i.e.,
$$||A||_2 = \sqrt{\lambda_\max (A^TA)}, $$ where $\lambda_\max (A^TA)$ is the maximum eigenvalue of $A^TA$. Solving
$$\mathrm{det}\: ( A^TA – \lambda \:I )=0$$ w.r.t. $\lambda$ we get the polynomial
$$p(\lambda) = \lambda^2 – ||A||_F^2\:\lambda + (\mathrm{det}\:A)^2=0$$ with discriminant
$$\Delta = ||A||_F^4 – 4(\mathrm{det} \:A)^2.$$ Thus, the maximum root of $p(\lambda)$ is
$$\lambda_\max(A^TA) = \frac{||A||_F^2 + \sqrt{||A||_F^4 – 4 (\mathrm{det}\:A)^2}}{2}$$ which is what the author in the book uses to get $||A||_2$, My question is, how do we know that $\Delta > 0$? Is there an inequality that gives an appropriate relation between $||A||_F$ and $\mathrm{det}\:A$ ? Could you please someone cast some light?
The author also has an exercise 2.8.8 where the square is missing in the determinant. Maybe it is a mistake.
Best Answer
For any $n \times n$ positive semidefinite matrix M we have
$$ \frac{\operatorname{tr}(M)}{n} \geq \operatorname{det}(M)^{1 / n} $$
And $A^T A$ is a special case of positive semidefinite matrix for any matrix $A$.
Since $\left(\operatorname{tr} A^{T} A\right)^{1 / 2} = \|A\|_{F}$ we have using the inequality above,
$$\|A\|_{F}^{4} = \left(\operatorname{tr} A^{T} A\right)^2 \geq n^2 (\det({A^TA}))^\frac{2}{n} = n^2 \det({A})^\frac{4}{n} $$
The special case for n=2 is given by :
$$\|A\|_{F}^{4} \geq 4 \det(A)^2$$
which is exactly what you want.
Of course there is possibly a much easier way to derive this inequality in the case of a $2 \times 2$ matrix !