# Eigenvalues of Sum of positive semi-definite matrices

Consider two real-valued positive semi-definite matrices $A$ and $B$, and suppose $C = A + B$. I am interested in proving that $\det(C) \geq \det(A)$.

I had heard through a colleague that the eigenvalues of $C$ are each bigger than either the eigenvalues of $A$ or the eigenvalues of $B$. Is there a name for this result?

If I could get a name for this result, I could argue that $\det(C) = \prod_{i=1}^n \lambda_i \geq \prod_{i=1}^n \gamma_i = \det(A)$, where $\gamma_i$ is the $i$'th eigenvalue of $A$ and $\lambda_i$ is the $i$'th eigenvalue of $C$.

Is there a name for this result, or a paper or book I could cite?

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If $A$ and $B$ are symmetric, this question math.stackexchange.com/questions/41463/… can help you (use a density argument). – Davide Giraudo Apr 28 '12 at 21:02
The matrices are symmetric, so I can apply that result directly. Nice. – Christopher Aden Apr 28 '12 at 21:07

## 2 Answers

We use the result given in this question: if $A$ and $B$ are two symetric positive definite matrices then $\det(A+B)>\det(A)$.

In your case, $A$ and $B$ are not necessary positive definite, but $A+B+2k^{-1}Id$ is so $\det(A+B+2k^{—1})> \det(A+k^{-1})$ and taking the limit $k\to +\infty$, $\det(A+B)\geq\det(A)$.

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@sevenkplus Thanks for editing. – Davide Giraudo Mar 7 '14 at 14:39

The following statement can be proved: given $A\succ 0$ and $B\succeq 0$, it holds that $\det(A+B)\geq\det(A)+\det(B)$, and equality holds iff $B=0$.

If one of your matrices is in fact positive definite (i.e., nonsingular), this answers your question. Is that of any help?

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