Professional music critiques rarely poses any aptitude for mathematics. As they like to compare musical processes unintelligible to them with the equally darksome methods of mathematical thinking, especially they hated algebra:

“The science of Mousier Berlioz is a sterile algebra” (P. Scudo, Critique et Littérature Musicales, Paris, 1852); “The music of Wagner imposes mental tortures that only algebra has a right to inflict” (Paul de Saint-Victor, La Presse, Paris, March 1861); “This theme serves [Ravel] for four movements [of the String Quartet] during which there is about as much emotional nuance as warms a problem in algebra” (New York Tribune, Dec 12, 1906).

Slonimsky, Lexicon of Musical Invective

Have they tried linear algebra? Things get better here.

Theorem 3 Let $A\in \R^{m\times r}$ be the left factor in an (economy) SVD, so its columns are orthonormal: $A^{\mathsf T}A=I_r$. Write $\tilde a_i\in\mathbb{R}^r$ for the $i$-th row of $A$, $|\tilde a_i|\le 1$.

Proof

Definition. A matrix $P$ is a projection matrix if (i) $P^2 = P$ (ii) $P$ is symmetric.

Theorem 1. If $P$ is a $m\times m$ projection matrix and $\rank{(P)} = r$, $P$ has $r$ eigenvalues equal to $1$ and $m - r$ equal to $0$. (Proof hint: $\forall v, \lambda^2 v = P^2v = Pv = \lambda v \Rightarrow \lambda = 1/0$).

Theorem 2. For any projection matrix, $0 \le P_{ii} \le 1, \forall i\in [m]$.

Proof.

Because $P = P^{\mathsf T} = P^2$, for any vector $x$, $$ x^{\mathsf T} P x = |P x|^2 ;\ge 0 . $$ Plug in $e_i$ (the standard basis vector with a 1 in the $i$-th entry), This guarantees that every diagonal entry is nonnegative: $$ P_{ii} = e_i^{\mathsf T} P e_i ;\ge 0 , \forall i\in [m] $$

Since $P$’s eigenvalues are either $0$ or $1$ (Theorem 1), its spectral norm is at most $1$. Equivalently: $$ x^{\mathsf T} P x ;\le; x^{\mathsf T} x \quad \text{for all } x . $$ Again, take $x = e_i$ (the standard basis vector with a 1 in the $i$-th entry). Then $$ P_{ii} = e_i^{\mathsf T} P e_i ;\le; e_i^{\mathsf T} e_i = 1 . $$

Let $P = AA^T$, so $|\tilde a_i|^2 = P_{ii} \in [0, 1]$. $P$ is a projection matrix cause $AA^TAA^T = AA^T$ (recall $A$ is column-orthogonal). Apply Theorem 2, the proof is complete.