Checking if $\det[\epsilon \mathbf{I} + \mathbf{A}] - \epsilon^n$ is a monotonic function

Question

Let $\mathbf{A}$ be an $n \times n$ positive semi-definite matrix, and let $\epsilon$ be a non-negative real number. I am interested in if the following function increases if $\epsilon$ increases (i.e., the function is monotonic with respect to $\epsilon$):

$$f(\epsilon) = \det[\epsilon \mathbf{I} + \mathbf{A}] - \epsilon^n $$

To check this, I can take the derivative of $f$ with respect to $\epsilon$ and check if the derivative is positive. The derivative is the following (assuming I did not make a mistake):

$$\frac{df}{d\epsilon} = \det[\epsilon \mathbf{I} + \mathbf{A}] \text{tr}[(\epsilon \mathbf{I} + \mathbf{A})^{-1}] - n\epsilon^{n-1}$$

However, I am having trouble showing this quantity is positive. I am fairly certain the quantity is positive based on some quick numerical experiments, but I want to verify analytically. Any thought on how to either show the first equation is monotonic or show the second equation is positive?

Answer 1

If $\lambda_1, \ldots, \lambda_n$ are the (necessarily real, non-negative) eigenvalues of $\mathbf{A}$ up to multiplicity, then $$\det[\varepsilon \mathbf{I} + \mathbf{A}] - \varepsilon^n = (\lambda_1 + \varepsilon)\ldots(\lambda_n + \varepsilon) - \varepsilon^n.$$ Indeed, we can dispense with determinants and the matrices, and simply consider the function $$f(\varepsilon) = (\lambda_1 + \varepsilon)\ldots(\lambda_n + \varepsilon) - \varepsilon^n$$ where $\lambda_1, \ldots, \lambda_n$ are arbitrary non-negative numbers. This is because we can form any positive-semidefinite $\mathbf{A}$ with these eigenvalues by simply placing them along the diagonal.

To prove $f$ is monotone, we can, as you tried, take the derivative. The derivative comes to, using the product rule, \begin{align*} f'(\varepsilon) &= \sum_{j=1}^n \prod_{\substack{1 \le i \le n \\ i \neq j}} (\lambda_i + \varepsilon) - n\varepsilon^{n-1} \\ &\ge \sum_{j=1}^n \prod_{\substack{1 \le i \le n \\ i \neq j}} \varepsilon - n\varepsilon^{n-1} \\ &= \sum_{j=1}^n \varepsilon^{n-1} - n\varepsilon^{n-1} \\ &= n\varepsilon^{n-1} - n\varepsilon^{n-1} = 0. \end{align*} Therefore, $f$ is monotone increasing.

Answer 2

Let $p(\lambda) = \det(\lambda I - A)$ be the characteristic polynomial. Then $$ \det(\varepsilon I + A) = (-1)^n\det (-\varepsilon I - A) = (-1)^np(-\varepsilon) $$ This is a monic polynomial of degree $n$ whose roots are minus the eigenvalues of $A$. When $A$ is positive semidefinite this means that the $n$ roots are all nonpositive. Therefore, Descartes' rule of signs tells us that there cannot be any sign changes among its coefficients. So all the coefficients are nonnegative, and this continues to hold when we subtract the leading term $\varepsilon^n$ away.

However, a polynomial with no negative coefficients is non-decreasing for positive arguments.

(If $A=0$ you do get $f(\varepsilon)=0$ which is strictly speaking not increasing -- but that special case probably won't bother you in practice).