If $A$ and $B$ are $n \times n$ symmetric matrices with eigenvalues bigger or equal with $0$, how can I prove that $\det(A+B) \geq \det A +\det B$?
How to prove that $\det(A+B) ≥ \det A +\det B$?
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matrices
inequality
eigenvalues-eigenvectors
determinant
positive-definite
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2See [this MathOverflow question](http://mathoverflow.net/questions/65424/determinant-of-sum-of-positive-definite-matrices). – 2017-01-07
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1This question seems somewhat similar: [How prove this matrix inequality $\det{(X)}\ge\det{(Y)}$](http://math.stackexchange.com/q/1101184) – 2017-01-07
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0i don't understand.. – 2017-01-07
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1The proof in [Andreas Thomas comment](http://mathoverflow.net/questions/65424/determinant-of-sum-of-positive-definite-matrices#comment435904_65430) in the aforementioned MO thread should be easy enough. – 2017-01-07
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2Possible duplicate of [Show that if $X \succeq Y$, then $\det{(X)}\ge\det{(Y)}$](https://math.stackexchange.com/questions/1101184/show-that-if-x-succeq-y-then-detx-ge-dety) – 2017-11-01