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Let $A$ and $B$ are two positive definite (by definition,automatically symmetric) matrices of order $n$.Now how can we show that $(det(A+B))^{\frac{1}{n}}\geq(detA)^\frac{1}{n}+(detB)^\frac{1}{n}$?

My attempt:

If $A$ and $B$ commute then we have a simultaneous diagonalization by an orthogonal matrix $P$.Then $\frac{LHS}{RHS}$ becomes a nice expression after substituting by $P$ and that diagonal matrix.Then $P$ gets cancelled and we can easily calculate those determinants since those are diagonal matrices.And the rest follows from basic Bernoulli's inequality.

Now we have to manage when $A$ and $B$ don't commute.Can anyone suggest me something? or I would be happy to learn if there is any other method to deal with it.

Thanks in advance.

Soumyadip Sarkar
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1 Answers1

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Every pair of positive definite matrices can be simultaneously diagonalised by congruence: let $A^{-1/2}BA^{-1/2}=QDQ^T$ be an orthogonal diagonalisation. Then $A=(A^{1/2}Q)(Q^TA^{1/2})$ and $B=(A^{1/2}Q)D(Q^TA^{1/2})$. Hence \begin{aligned} &(\det(A+B))^{1/n}-\det(A)^{1/n}-\det(B)^{1/n}\\ =\,&\det(A^{1/2}Q)^{1/n}\left[\det(I+D)^{1/n} -1-\det(D)^{1/n}\right]\det(Q^TA^{1/2})^{1/n}\\ =\,&\det(A)^{1/n}\left[\det(I+D)^{1/n} -1-\det(D)^{1/n}\right] \end{aligned} and you may continue from here.

user1551
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  • Hello!! I have recently learnt that,you don't need both of them to be PD,just at least one of them would suffice. – Soumyadip Sarkar Jul 24 '19 at 15:37
  • @SoumyadipSarkar While you don't need both of them to be PD in order to simultaneously diagonalised by congruence, you do need them to be PD to guarantee that $(\det A)^{1/n}$ and $(\det B)^{1/n}$ are real numbers. – user1551 Jul 24 '19 at 15:40
  • No,i mean for simultaneous diagonalization you don't both of them,i am not talking about my question. – Soumyadip Sarkar Jul 24 '19 at 17:50