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.
Best Answer
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.