My goal is to show that :

$$\forall A,B\in O_n(\mathbb{R}), |{\rm det}(A+B)| \le 2^n.$$

How can I access the determinant of the sum of two matrices ? I don’t know many ways to establish inequalities in Algebra, except with Bessel’s inequality and Cauchy Schwarz inequality, but they don’t seem to be very helpful here.

  • $\begingroup$ Without loss of generality you may assume one of the two matrices $A,B$ is the identity. $\endgroup$ – hardmath Jan 18 at 23:07

Since $AA^T = I$, we have $$\det (A+B)= \det A(I+A^TB) = \det A \,\det(I+A^TB).$$ Now, $|\det A| = 1$ since $A$ is orthogonal, and $A^TB$ is orthogonal as well so all its eigenvalues $\lambda_1, \ldots, \lambda_n$ are on the unit sphere in $\Bbb{C}$. Therefore $$|\det(A+B)| = |\det(I+A^TB)| = |(1+\lambda_1)\cdots(1+\lambda_n)| \le (1+|\lambda_1|)\cdots(1+|\lambda_n|) \le 2^n.$$


Mechanodroid's solution is great. Here's a slightly different way to look at it.

Suppose $\lambda$ is an eigenvalue of $A+B$ with corresponding unit-length eigenvector $v$. Then,

$$|\lambda| = \|\lambda v\| = \|(A+B)v\| = \|Av+Bv\| \le \|Av\|+\|Bv\| = \|v\|+\|v\| = 2\|v\| = 2,$$

where we have used the fact that $\|Av\| = \|v\|$ and $\|Bv\| = \|v\|$ since $A,B$ are orthogonal. Since all $n$ eigenvalues of $A+B$ have magnitude at most $2$, the determinant of $A+B$ which is the product of the $n$ eigenvalues of $A+B$ is at most $|\det(A+B)| \le 2^n$.


another proof, no eigenvalues required:
since this is over $\mathbb R$, instead of working with absolute values, work with squares

i.e. prove ${\rm det}(A+B)^2\leq 2^{2n}$

special case: consider if both matrices are the identity, then
$\big \vert{\rm det}(I+I)|^2 = \det\Big(\big (I+I)^T\big (I+I)\Big) = \det\big(I+2I+I\big)=\prod_{k=1}^n (1+1+1+1) = 2^{2n}$

in general:
$\det\big((A+B)^2\big) $
$= \det\Big(\big (A+B)^T\big (A+B)\Big) $
$= \det\big(I+A^TB + B^TA+I\big)$
$\leq \prod_{k=1}^n (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)$
$=\prod_{k=1}^n \big \vert (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)\big \vert$
$\leq \prod_{k=1}^n \Big(1+\big \vert (A^TB)_{k,k}\big \vert+\big \vert(B^T A)_{k,k}\big \vert+1\Big)$
$\leq \prod_{k=1}^n \Big(1+1+1+1\Big)$
$= 2^{2n}$
1.) Hadarmard Determinant Inequality
2.) any matrix of the form $C^T C$ is PSD so it has real non-negative diagonals
3.) Triangle inequality (and multiplying over a point-wise bound)
4.) $(A^TB)$ and $(B^TA)$ are real orthogonal matrices so each component has modulus $\leq 1$

  • $\begingroup$ I don't understand your justification of $\det\big(I+A^TB + B^TA+I\big) \leq \prod_{k=1}^n (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)$ by Hadamard determinant inequality (en.wikipedia.org/wiki/Hadamard%27s_inequality) in particular why solely diagonal entries are kept ? $\endgroup$ – Jean Marie Jan 22 at 14:52
  • $\begingroup$ @JeanMarie over reals, we have $C:= A+B$, then $G:= C^TC$ is real symmetric PSD, and $\det\big(C^TC\big)=\det\big(G\big) \leq \prod_{k=1}^n g_{k,k}$.-- i.e. the product of the diagonals gives an upper bound on the determinant of any Gram matrix -- this is Hadamard's Determinant Inequality. It's provable purely via Cholesky Decomposition which I just used to prove a refinement of it here: math.stackexchange.com/questions/3992474/… . Higher brow interpretation: the diagonal of HPSD matrix is majorized by the eigenvalues and determinant is Schur Concave. $\endgroup$ – user8675309 Jan 22 at 19:18
  • 1
    $\begingroup$ I see! I wasnt aware of this Hadamard inequality. Thank you very much for your time. $\endgroup$ – Jean Marie Jan 22 at 19:24

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