7
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
6
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.