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Let $A,B\in M_{n}(\mathbb{R})$ so that $A>0, B>0$.

I need to prove that $\det (A+B)>\max (\det(A), \det(B))$.

I want to use Sylvester theorem of having a matrix $D$ so that $D=\operatorname{diag}(1,1,\ldots, 1,-1,-1, \ldots-1,0,0,\ldots,0)$.

Do I need to use it? How do I use it here?

Thank you.

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It means that A is symmetric and all it's eigenvalues is bigger than 0. –  user6163 May 26 '11 at 12:27

2 Answers 2

up vote 1 down vote accepted

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.

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Thank you for the comment, Can you extend 2? –  user6163 May 26 '11 at 22:31
    
Have you learned the theorem which states that every quadratic form in an euclidian space has an orthonormal basis where its matrix is diagonal? It is linked to the sepctral theorem. Try to look at this (first paragraph after Properties) en.wikipedia.org/wiki/Symmetric_matrix –  user10676 May 27 '11 at 8:09
    
Yes , I learned that, I just can't understand what I do with that in this question. –  user6163 May 27 '11 at 11:23

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 $\det(PA^tP)=\det A$.

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Very nice. I see you used $B$ symmetric. If $B>0$ then $B$ must be symmetric? –  Beni Bogosel May 26 '11 at 11:05

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