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.

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    $\begingroup$ 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. $\endgroup$ – Aleksei Kulikov Dec 24 '18 at 1:11
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    $\begingroup$ 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 ($|A|=(A^*A)^{1/2}$). $\endgroup$ – 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.

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    $\begingroup$ we could use Hadamard inequality $|\det A|\leqslant \prod_{i=1}^n \|Ae_i\|\leqslant (\frac1n \sum \|Ae_i\|^2)^{n/2}=n^{-n/2} \|A\|_{HS}^n$ $\endgroup$ – 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)


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