$\det(I - A) \ge \prod_{i} [1 - \sigma_i(A)]$ if $\|A\|_2\le 1$

Question

I would like to prove $$ \det(I - A) \ge \prod_{i=1}^n[1-\sigma_i(A)] $$ for any $A\in\mathbb{R}^{n\times n}$ such that $\|A\|_2 \le 1$. Here, as usual, $\sigma_i(\cdot)$ denotes the $i$-th sigular value.

Here is a proof, but I am looking for one with less computation.

Proof. Note that $\det(I-A)\ge 0$. This is because the eigenvalues of $I-A$ are either nonnegative or come in conjugate pairs. Therefore, it suffices to prove that $$ [\det(I-A)]^2 \ge \prod_{i=1}^n[1-\sigma_i(A)]^2. $$ We focus on the case where $\|A\|_2<1$. Let the SVD of $A$ be $$ A = U\Sigma V^*. $$ Denote $W=U^*V$. It is easy to verify that $$|\det(I-A)| = |\det(W-\Sigma)|.$$ Thus we only need to prove $$ \det(W-\Sigma)\det(W^*-\Sigma) \ge [\det(I-\Sigma)]^2. $$ This can be done by observing that $$ (W-\Sigma)(I-\Sigma)^{-1}(W^*-\Sigma) \succeq I -\Sigma, $$ which holds because $$ \begin{split} \;&(W-\Sigma)(I-\Sigma)^{-1}(W^*-\Sigma) - (I-\Sigma)\\ =\;&W(I-\Sigma)^{-1}W^* + \Sigma(I-\Sigma)^{-1}\Sigma - W(I-\Sigma)^{-1}\Sigma - \Sigma(I-\Sigma)^{-1} W^* - (I -\Sigma)\\ =\;&W[(I-\Sigma)^{-1}-I]W^* + [\Sigma(I-\Sigma)^{-1}\Sigma + \Sigma] - W(I-\Sigma)^{-1}\Sigma - \Sigma(I-\Sigma)^{-1} W^*\\ =\;& W[\Sigma(I-\Sigma)^{-1}]W^* + \Sigma(I-\Sigma)^{-1} - W[\Sigma(I-\Sigma)^{-1}] - [\Sigma(I-\Sigma)^{-1}] W^*\\ =\;&(W-I)[\Sigma(I-\Sigma)^{-1}](W^*-I)\\ \succeq\; & 0. \end{split} $$ It is helpful to note that $I-\Sigma$ is a positive definite diagonal matrix. Q.E.D.

The last step of this proof is straightforward but needs some computation. Does there exist a proof that is "simpler" in terms of computation, maybe resorting to some other inequalities?

The major technique of this proof comes from Sec. 3.8 of [S. G. Wang, M. X. Wu, and Z. Z. Jia, Matrix Inequalities (in Chinese), Science Press, Beijing, 2005], where the same skill proves Hua's inequality, namely $$ \det(I-AA^*)\det(I-BB^*) \le |\det(I-AB^*)|^2 $$ for any complex matrices $A$ and $B$ of the same size and $AA^* \preceq I$, $BB^* \preceq I$. Hua proved the inequality in 1955. It has applications in multivariate complex analysis. Indeed, the desired inequality is a corollary of Hua's inequality, which tells us $$ |\det(I-A)|^2 \ge \det(I - U\Sigma U^*)\det(I - V\Sigma V^*) = [\det(I-\Sigma)]^2. $$ But Hua's inequality may not be familiar enough to be quoted here.

Any comments or criticism will be appreciated. Thank you very much.

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As suggested by @Ben Grossmann, here is a nice approach by Weyl's inequality for singular values.

Theorem ([Thereom 3.3.16, R. A. Horn and Ch. R. Johnson, Topics on Matrix Theory, Cambridge University Press, 1991]).
Let $A$, $B\in\mathbb{C}^{m\times n}$ be given, and let $q=\min\{m,n\}$. The following inequalities hold for the decreasingly ordered singular values.

1. $\sigma_{i+j-1}(A+B)\le \sigma_i(A)+\sigma_{j}(B)$,
2. $\sigma_{i+j-1}(AB^*)\le \sigma_i(A)\sigma_{j}(B)$,

where $1\le i, j\le q$ and $i+j \le q+1$.

Here we need only the first inequality, which is known as Weyl's inequality (what about the second one?).

The following proof is a copy of @Ben Grossmann's answer.

Let us order the singlar values decreasingly. Then Weyl‘s inequality leads us to
$$\sigma_{i}(I-A)+\sigma_{n-i+1}(A) \ge \sigma_n(I) = 1 \quad \text{for each} \quad i=1,2,..., n. $$ Note that $\det(I-A)\ge 0$ as in the aforementioned proof. Therefore, $$ \det(I-A) = |\det(I-A)| =\prod_{i=1}^n[\sigma_i(I-A)]\ge \prod_{i=n}^1[1-\sigma_i(A)]= \prod_{i=1}^n[1-\sigma_i(A)]. $$

Answer 1

I claim that $\sigma_i(I - A) \geq 1 - \sigma_{n+1-i}(A)$. We could prove this, for example, using the min-max theorem: $$ \sigma_i(I - A) = \min_{\dim(S) = n-i+1} \max_{x \in S, \|x\| = 1} \|(I - A)x\|\\ = \min_{\dim(S) = n-i+1} \max_{x \in S, \|x\| = 1} \|x - Ax\|\\ \geq \min_{\dim(S) = n-i+1} \max_{x \in S, \|x\| = 1} (\|x\| - \|Ax\|)\\ = \min_{\dim(S) = n-i+1} \max_{x \in S, \|x\| = 1} (1 - \|Ax\|)\\ = 1 - \max_{\dim(S) = n-i+1} \min_{x \in S, \|x\| = 1} \|Ax\|\\ = 1 - \sigma_{n+1-i}(A). $$ It follows that $$ |\det(I - A)| = \prod_i \sigma_i(I - A) \geq \prod_i [1 - \sigma_{n + 1 - i}(A)] = \prod_i [1 - \sigma_i(A)]. $$ Finally, $\det(I - A)\ge 0$ because the eigenvalues of $I-A$ are either nonnegative or come in conjugate pairs.

Answer 2

For a simple proof using weak Majorization:

1.) consider that if $\big \Vert A\big \Vert_2 =1$ there is nothing to do as the RHS is zero and the LHS is necessarily real non-negative. So consider $\big \Vert A\big \Vert_2 \lt 1$

2.) It suffices to check the case where where each $\lambda_i\geq 0$ -- otherwise use the positivity of the determinant and triangle inequality to show any other result for the LHS is bounded below by the real non-negative eigenvalues case. Now recall that the singular values of a matrix weakly majorize the modulus of the eigenvalues of said matrix.

3.) consider for $x_i\in[0,1)$: the function $f$ given by $f(x_i) = -\log(1-x_i)$. With this domain, $f$ is strictly convex and increasing (check 1st and 2nd derivatives).

4.) Using the symmetric function $g(\mathbf x\big) = \sum_{i=1}^n f(x_i)$ which is Schur convex we have
$g\left(\begin{bmatrix}\sigma_1\\\vdots \\\sigma_n\end{bmatrix}\right)\geq g\left(\begin{bmatrix}\lambda_1\\\vdots \\\lambda_n \end{bmatrix}\right)$
negating each side this reads

$\log(1-\lambda_1)+...+\log(1-\lambda_n)\geq \log(1-\sigma_1)+...+\log(1-\sigma_n)$
or

$\det(I - A) \ge \prod_{i=1}^n[1-\sigma_i(A)]$