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

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.

• This is essentially what you ask: "Let $p(x)\in\mathbb{C}[x]$ be a monic polynomial of degree $n$. I know the value of $p(t)$ for some $t\in\mathbb{C}$. What do we know about $p(0)$?" The answer is nothing. For any $t\neq 0$ and any possible value of $p(t)$, $p(0)$ can take any value, provided that $n>1$. Jul 6, 2016 at 14:59
• @Batominovski, thank you for the reply. How is det(A + X), det(X) related to the polynomial of p(t) , p(0) in your answer ? Jul 6, 2016 at 15:25
• @david Take $p(x):=\det(x\,I-X)$ (i.e., the characteristic polynomial of $X$), and you will see. Jul 6, 2016 at 15:48
• @david The situation is rather specific to the case where the shift is by a multiple of $I$. In this case the function $p(\lambda)=\operatorname{det}(A+\lambda I)$ is a polynomial. In general it can be any monic polynomial of degree $n$ (there is a so-called "companion matrix" for any monic polynomial, which has that polynomial as its characteristic polynomial). You're saying that you know $p(1)$ and want to know $p(0)$, but this cannot happen.
– Ian
Jul 6, 2016 at 15:49

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 $\bf X\neq I$.

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 $\det(I+X) > \det(X)$.

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,

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

• The OP wants $\det(X+I)$, not $\det(X+1_n1_n^T)$. Jul 6, 2016 at 23:05
• @egreg Please read the 2nd paragraph. Jul 6, 2016 at 23:06
• “In general…” doesn't mean “for a carefully selected case”. ;-) Jul 6, 2016 at 23:07
• @egreg The OP wrote "constant matrix". The case is not carefully selected, it's what he asked for. Jul 6, 2016 at 23:09
• @RodrigodeAzevedo, thank you for the reply. I am not familiar with the symbol you used. What is a1n, 1Tn in your expression ? Jul 7, 2016 at 1:30

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: $$\| \det(A+X) -\det(A) \| \leq C(d) \|X\| \max(\|A+X\|^{d-1},\|A\|^{d-1})$$