# An upper estimate for $|\det(A+B)|$

If $$A$$ and $$B$$ are $$n\times n$$ matrices, then it easily follow from the definition of the determinant by sum over permutations, and from the Young inequality that $$|\det (A+B)|\leq C(n)(\Vert A\Vert^n+\Vert B\Vert^n),$$ where $$\Vert \cdot\Vert$$ stands for the Hilbert-Schmidt norm of a matrix. I am looking for stronger/better/more elegant estimates (with references included) that would imply the above inequality.

• Are you interested in some generalizations of this inequality or best possible $C(n)$? I have reasonably short proof that $C(n) = \frac{2^{n-1}}{\sqrt{n}^n}$ (that is, $A = B = I$ is optimal), but if you want something else could you please specify it somehow. – Aleksei Kulikov Dec 24 '18 at 1:11
• The bound $|\det(A+B)|=\det|A+B| \le \prod_i (a_i+b_{n-i+1})$, where $a_i$ and $b_i$ are the singular values of $A, B$, respectively (sorted in the usual decreasing order) easily holds. Moreover, it follows from the following even more elegant inequality: $|\det(A+B)|^2 \le \det(|A|+|B|)\det(|A^*|+|B^*|)$, where $|A|$ denotes the matrix absolute value ($|A|=(A^*A)^{1/2}$). – Suvrit Dec 24 '18 at 12:51

Okay, here is the proof that $$C(n) = \frac{2^{n-1}}{\sqrt{n}^n}$$.

Firstly, it is enough to find best constant $$c(n)$$ in $$|\det(A)| \le c(n)||A||^n$$. Indeed, we have $$|\det(A+B)| \le c(n)||A+B||^n \le c(n)(||A|| + ||B||)^n \le 2^{n-1}c(n) (||A||^n + ||B||^n),$$ and for $$A = B$$ we have here equality.

Now let $$A$$ be a matrix with $$||A|| = 1$$ and maximum possible $$|\det A|$$ (such a matrix obviously exists by compactness). Let $$A = (v_1, \ldots , v_n), v_k\in \mathbb{C}^n$$. We will prove that $$(v_k, v_m) = 0, k\ne m$$ and $$(v_k, v_k) = (v_m, v_m)$$.

Assume that $$(v_k, v_m) \ne 0, k\ne m$$. Then we can add $$tv_k$$ to $$v_m$$ without changing determinant. But if $$(v_k, v_m) \ne 0$$ then we can choose $$t$$ such that $$(v_m + tv_k, v_m + tv_k) < (v_m, v_m)$$. But then we will have matrix with smaller Hilbert-Schmidt norm and the same $$|\det|$$, which is impossible since $$A$$ gives us maximum.

Similarly assume that $$(v_k, v_k) \ne (v_m, v_m)$$. Then we can instead consider $$\lambda v_k$$, $$\frac{1}{\lambda} v_m$$ without changing $$\det$$. But then for $$\lambda$$ close enough to $$1$$ we can again decrease norm of $$A$$ which is forbidden.

Thus, all columns of $$A$$ are orthogonal and have same length. Then $$A$$ is a multiple of unitary matrix and since $$\det$$ of all unitary matrices is $$1$$ and Hilbert-Schmidt norm of all unitary matrices is $$\sqrt{n}$$ we are done.

• we could use Hadamard inequality $|\det A|\leqslant \prod_{i=1}^n \|Ae_i\|\leqslant (\frac1n \sum \|Ae_i\|^2)^{n/2}=n^{-n/2} \|A\|_{HS}^n$ – Fedor Petrov Dec 24 '18 at 9:23

There is a paper by Chi-Kwong Li and Roy Mathias about lower and upper bounds for $$|\det(A + B)|$$ in terms of the singular values of $$A$$ and $$B$$. (The determinant of the sum of two matrices)