$A,B$ are two real positive matrices then $\det (A+B) > \max(\det A , \det B)$

Question

Let $A,B$ two square-real-positive matrices. Prove that $\det (A+B) > \max(\det A , \det B)$

So I found this solution: https://math.stackexchange.com/a/41478/160028

Basically, if $A=I_n$ and $B$ is diagonal then the proof is immediate.

Now, I know that if $M$ is positive-symmetric then:

• $M$ is conjugate to $I_n$.
• $M$ is conjugate to $\text{Diag}(c_1,\ldots,c_n)$ where $c_i > 0$.

but as far as I understand it doesn't have to be the same $P$. Anyhow, how do I utilize it in order to prove the inequality?

Thanks.

Answer 1

We can show that a satisfactory $P$ exists as follows:

First, suppose we only have $P^TAP = I$. Then apply the spectral theorem to find an orthogonal $U$ such that $U^T(P^TBP)U = D$ is diagonal. We then have $$ (PU)^T(A + B)(PU) = \\ U^T(P^TAP)U + U^T(P^TBP)U = \\ U^T(I)U + D =\\ I + D $$ So, the new matrix $PU$ does the job.