Why $\sigma_i(A)^2 |a_i|^2 + \dots + \sigma_q(A)^2 |a_q|^2 \leq \sigma_i(A)^2$ holds?

Question

Let $A \in \mathbb{R}^{m \times n}$, $q = \min {m, n}$ with singular values $\sigma_i(A) \geq \dots \geq \sigma_q(A)$ and $a_i^2 + \dots +a_q^2 = 1$. Could you please someone help to understand why the following inequality holds?

$$\sigma_i(A)^2 |a_i|^2 + \dots + \sigma_q(A)^2 |a_q|^2 \leq \sigma_i(A)^2$$

^{This is part of the proof of Lemma 3.3.15 in Topics in Matrix Analysis,Roger A. Horn.}

Answer 1

You have that:

$$ \sigma_{j}(A) \leq \sigma_{i}(A) , \forall j \in [i,q]$$

And as the singular values are non-negative:

$$ \sigma_{j}(A)^2 \leq \sigma_{i}(A)^2 , \forall j \in [i,q]$$

Thus the left hand side is less than:

$$ \sigma_{i}(A)^2 (a_{i}^2 + \cdots + a_{q}^2) = \sigma_{i}(A)^2 $$