8
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Without loss of generality you may assume one of the two matrices $A,B$ is the identity. $\endgroup$ – hardmath Jan 18 at 23:07
15
$\begingroup$

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 $$|\det(A+B)| = |\det(I+A^TB)| = |(1+\lambda_1)\cdots(1+\lambda_n)| \le (1+|\lambda_1|)\cdots(1+|\lambda_n|) \le 2^n.$$

$\endgroup$
9
$\begingroup$

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 $|\det(A+B)| \le 2^n$.

$\endgroup$
1
$\begingroup$

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) $
$= \det\big(I+A^TB + B^TA+I\big)$
$\leq \prod_{k=1}^n (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)$
$=\prod_{k=1}^n \big \vert (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)\big \vert$
$\leq \prod_{k=1}^n \Big(1+\big \vert (A^TB)_{k,k}\big \vert+\big \vert(B^T A)_{k,k}\big \vert+1\Big)$
$\leq \prod_{k=1}^n \Big(1+1+1+1\Big)$
$= 2^{2n}$
justification:
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 $\leq 1$

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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