How much can we tell about $\det(X)$ if we know $\det(I + X)$?

Question

1. What can we tell about $\det(X)$ if we know $\det(I + X)$? Will it give some kind of bound for $\det(X)$?
2. In general, if we know the determinant of matrix $A + X$, where $A$ is a constant matrix, how much can we say about $\det(X)$?

Thank you for the attention.

Answer 1

The determinant is the product of the eigenvalues $$\det({\bf A}) = \prod_k \lambda_k({\bf A})$$

Then it is well known that all eigenvalues increase by 1 if adding $\bf I$ to any matrix:

$$\det({\bf A+I}) = \prod_k (\lambda_k({\bf A})+1)$$

Maybe you can rewrite this product into something you are comfortable working with. I do not think this extends in any nice way to adding $\bf X\neq I$.

Answer 2

Let $\{\lambda_1,\dots,\lambda_n\}$ be the set of eigenvalues of $X$. Then the set of eigevalues of $I+X$ is $\{1+\lambda_1,\dots,1+\lambda_n\}$. Determinant of any matrix $X$ is equal to product of all eigenvalues of $X$, thus \begin{align} \det(X) = \prod_{i=1}^n\lambda_i &&\det(I+X) = \prod_{i=1}^n(1+\lambda_i) \tag{1} \end{align} From (1) it is clear that (in general) nothing can be said about $\det(I+X)$.

If all eigenvalues of $X$ are real and positive (i.e. $X$ is positive definite), then $\det(I+X) > \det(X)$.

Answer 3

If we are given an invertible matrix $X \in \mathbb R^{n \times n}$ and the following determinant

$$\beta := \det (X + \alpha 1_n 1_n^T)$$

then

$$\beta = \det (X + \alpha 1_n 1_n^T) = \det (X) \cdot \underbrace{\det (I_n + \alpha X^{-1} 1_n 1_n^T)}_{= 1 + \alpha 1_n^T X^{-1} 1_n} = (1 + \alpha 1_n^T X^{-1} 1_n) \cdot \det (X)$$

where we used Weinstein-Aronszajn determinant identity. Thus,

$$\det (X) = \frac{\beta}{1 + \alpha 1_n^T X^{-1} 1_n}$$

Answer 4

Nothing To document that "nothing if $d>1$" where $d$ is the dimension is the answer (as said above) :

Claim: For any $\lambda\in \mathbb R$ there exists a matrix $X$ such that $\det X = \lambda$, while $\det(I+X)=0$.

As a corollary (take $\lambda = \pm n\in\mathbb N$ in the first claim),

Corollary: there exists a sequence $X_n$ such that $\det X_n \to \pm \infty$ while $\det(I+X_n)=0$.

This tells you that nothing general can be said.

Proof. If $d$ is odd take $X$ to be be diagonal matrix with $X_{11}=\lambda$ and $X_{kk}=1$ for $2\leq k\leq d-1$ and $X_{dd}=-1$. If $d$ is even and greater or equal to $4$, $X_{kk}=1$ for $2\leq k\leq d-2$ and $X_{d-1,d-1}= X_{dd}=-1$. if$d=2$, take $X=\begin{pmatrix} -1 & \lambda \\ -1 & -1\end{pmatrix}$.

Something

On the other hand, it could be that you meant something else than what you asked. For example, what is true is Hadamard's inequality: $$ \| \det(A+X) -\det(A) \| \leq C(d) \|X\| \max(\|A+X\|^{d-1},\|A\|^{d-1}) $$