# Determinant of sum of positive definite matrices

Say $A$ and $B$ are symmetric, positive definite matrices.

I've proved that $\det(A+B) \ge \det(A) + \det(B)$ in the case that $A$ and $B$ are two dimensional.

Is this true in general for $n$-dimensional matrices?

Is it even true that $\det(A+B) \ge \det(A)$ [as this would also be enough..]

Thanks.

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## 6 Answers

The inequality $$\det(A+B)\geq \det A +\det B$$ is implied by the Minkowski determinant theorem $$(\det(A+B))^{1/n}\geq (\det A)^{1/n}+(\det B)^{1/n}$$ which holds true for any non-negative $n\times n$ Hermitian matrices $A$ and $B$. The latter inequality is equivalent to the fact that the function $A\mapsto(\det A )^{1/n}$ is concave on the set of $n\times n$ non-negative Hermitian matrices (see e.g., A Survey of Matrix Theory and Matrix Inequalities by Marcus and Minc, Dover, 1992, P. 115 and also the previous MO thread).

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looks nice, but isn't it somewhat overkill to invoke the Minkowski theorem here? –  Suvrit May 19 '11 at 17:05
Thanks very much! –  user15221 May 19 '11 at 17:44
This is really misleading since it makes the question look like complicated, while is almost obvious that $\det(A+B) = \det(A) \det(1+A^{-1/2}BA^{-1/2}) \geq \det(A) (1+\det(A^{-1/2}BA^{-1/2})) = \det(A) + \det(B)$. In the second step, it is just used that $\prod_i (1+ \mu_i) \geq 1 + \prod_i \mu_i$, where $\mu_i$ are the eigenvalues of $A^{-1/2}BA^{-1/2}$. –  Andreas Thom Jun 28 at 20:02

Yet another way to see this is to note that $A = \overline{Q}^{t}Q$ for some invertible matrix $Q$. Then ${\rm det}(A+B) = |{\rm det}(Q)|^{2}{\rm det}{( I + (\overline{Q}^{-1}})^{t}BQ^{-1})$.` Now $(\overline{Q}^{-1})^{t}BQ^{-1}$ is Hermitian, and positive definite. It suffices to prove that if $X$ is positive definite and Hermitian, then ${\rm det}(I+X) \geq (1 + {\rm det}X)$. We may conjugate $X$ by a unitary matrix $U$ and assume that $X$ is diagonal. Let the eigenvalues of $X$ be $\lambda_{1},\ldots, \lambda_{n}$, (allowing repetitions). Then ${\rm det}(I+X) = \prod_{i=1}^{n}(1 + \lambda_{i}) \geq 1 + \prod_{i=1}^{n} \lambda_{i} = 1 + {\rm det}X.$ Such an argument appears in some proofs by R. Brauer, though I do not know whether it originates with him.

Later edit: Incidentally, I think that with the arithmetic-geometric mean inequality and a slightly more careful analysis, you can see by this approach that for $X$ as above, you do have ${\rm det}(I+X) \geq (1 +({\rm det}X)^{1/n})^{n}$ (a special case of the inequality of Minkowski mentioned in the accepted answer, but enough to prove the general case by an argument similar to that above). For set $d = {\rm det}X$. Let $s_{m}(\lambda_{1},\ldots ,\lambda_{n})$ denote the $m$-th elementary symmetric function evaluated at the eigenvalues. Using the arithmetic-geometric mean inequality yields that $s_{m}(\lambda_{1},\ldots ,\lambda_{n}) \geq \left( \begin{array}{clcr} n\\m \end{array} \right)d^{m/n}$, so we obtain ${\rm det}(I+X) \geq (1+d^{1/n})^{n}.$

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We have $((A+B)x,x)\ge (Ax,x)$. It then follows from the variational characterization of eigenvalues (min-max theorem) that the eigenvalues of $A+B$ are greater than or equal to those of $A$. This implies $det(A+B)\ge det(A)$.

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Here is yet another overkill, but hopefully not too bad a way to prove this inequality.

We have the following proof sketch.

$$\begin{eqnarray} x^T(A+B)x &\ge& x^TAx\quad\forall x\\\\ -x^T(A+B)x &\le& -x^TAx\\\\ \exp(-x^T(A+B)x) &\le& \exp(-x^TAx)\\\\ \int\exp(-x^T(A+B)x)dx &\le& \int\exp(-x^TAx)dx\\\\ \frac{1}{\sqrt{\det(A+B)}} &\le& \frac{1}{\sqrt{\det(A)}}\\\\ \det(A+B) &\ge& \det(A) \end{eqnarray}$$

The only fancy thing that happened is in the second last line, where I used the formula for the Gaussian integral (see multivariate section)

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The determinant of a positive definite matrix $G$ is proportional to $(1/\hbox{Volume}(\mathcal B(G)))^2$ where $\mathcal B(G)$ denotes the unit ball with respect to the metric defined by $G$. If $A$ and $B$ are positive definite then the volume of $\mathcal B(A+B)$ is smaller than the volume of $\mathcal B(A)$ or $\mathcal B(B)$.

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It's worth noting that this is secretly the same as Suvrit's answer. –  Mark Meckes May 20 '11 at 14:20
Not really: You don't need exponentials for proving that $\det(G)$ is proportional to $1/\hbox{Volume}(G)^2$ : It is enough to stare at an orthogonal basis formed of eigenvectors for $G$. In this sense this proof is more elementary. –  Roland Bacher May 25 '11 at 7:18
Fair enough. –  Mark Meckes May 29 '11 at 0:49

Let me add some more. If $A, B, C$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C). \quad (\star)$$

When $C=0$, this reduces to OP's question.

A remarkable extension of ($\star$) were recently obtained by V. Paksoy, R. Turkmen, F. Zhang [ Electron. J. Linear Algebra 27 (2014) 332-341], which says that the determinant functional can be replaced by any generalized matrix function.

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