$A,B\in M_{n}(\mathbb{R})$ so that $A>0, B>0$, prove that $\det (A+B)>\max (\det(A), \det(B))$

Solution 1:

1) First prove your result when $A=I_n$ and $B$ is diagonal.

2) Next, use the theorem of diagonalisation of quadratic forms : If $A>0$ and $B$ is symmetric, then there exists $P$ invertible such that $P^{t}AP=I_n$ and $P^{t}BP$ is diagonal.

Solution 2:

We begin by the case in which for example $B$ is diagonal, which diagonal coefficients $b_1,\ldots,b_n$. If we expand the determinant $\det(A+B)$ thank to multi-linearity, we will get a sum of $2^n$ determinants. One is $\det A$ (if we choose the column of $A$) an one is $\det B$. In the others, if we expand the determinant according to the column which correspond to column of $B$, we get the determinant of a matrix related to $A$. For example, if $1\leq i_1<\ldots<i_k\leq n$ are the indices of these column, the determinant is $\prod_{j=1}^kb_{i_j}\times$ the determinant of the matrix $A'$ which is $A$ but without the lines and columns $i_1,\ldots,i_k$. This matrix is positive because if $x'=(x_1,\ldots,x_k)$ then if we put $x\in\mathbb{R}^n$ such that $x_j = x'_j$ if $j\in\left\{i_1,\cdots,i_k\right\}$ and $x_j=0$ otherwise then $^txAx = ^tx'A'x'$. We get $\det(A+B)\geq \det A+\det B$. In the general case, since $B$ is symmetric, we can find $P$ such that $^tPP=I$ and $B =^tPDP$ with $D$ diagonal. We apply the previous result to $PA^tP$ and $D$. We can conclude since $\det(PA^tP+D) =\det(PA^tP+PB^tP)=\det (A+B)$ and $\det(PA^tP)=\det A$.