The following theorem is called the norm-trace inequality. This ineqality plays a major role in the Minkowski-Siegel reduction theory in [1], and therefore in [2]
To understand the setup here, refer to the page on real semisimple algebras
Consider a finite-dimensional semisimple \(\mathbb{R}\)-algebra \(A\) with a positive involution \((\ )^{*}\). Let \(a \in A\) be a symmetric positive definite element and let \(d = \dim_{\mathbb{R}} A\). Then \(\mathop{\mathrm{N}}(a) > 0\), \(\mathop{\mathrm{tr}}(a) > 0\) and \[\begin{align} \frac{1}{d}\mathop{\mathrm{tr}}(a) \ge \mathop{\mathrm{N}}(a)^{\frac{1}{d}}. \end{align}\]
This is just the arithmetic-geometric means inequality. Let us elaborate how.
We know that \(x \mapsto \mathop{\mathrm{tr}}(x^* y)\) is an inner product on \(A\). With respect to this, construct an orthonormal basis \(e_1, e_2 ,\dots, e_d\). Set \(a_{ij} = \mathop{\mathrm{tr}}(e_i^{*} a e_{j})\) which are the matrix entries of left-multiplication by \(a\) with respect to the basis \(\{ e_{i}\}_{i=1}^{d}\), i.e. for \(\{ r_i\}_{i=1}^{d} \subseteq \mathbb{R}^{d}\), \(a ( \sum_{i} r_i e_i) = \sum_{i}^{} ( \sum_{j} a_{ij}r_j) e_i\). Since \(a\) is symmetric, we get that \(a_{ij} = a_{ji}\). Furthermore, by the positive definiteness of \(x \mapsto \mathop{\mathrm{tr}}(x^{*} a x)\), the matrix \(a_{ij}\) can be seen to be positive definite as a real matrix by substituting \(x = \sum_{i=1}^{d} x_i e_i\).
Hence using the spectral theorem for real positive definite symmetric matrices, \(a_{ij}\) is diagonalizable matrix with respect to an orthonormal change of basis and has real and positive eigenvalues (i.e. the diagonal entries). Then trace is the sum of those eigenvalues and the norm is the product. The inequality is then exactly the arithmetic-geometric inequality on those eigenvalues.
This page was updated on October 11, 2021.
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