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 the following even true?

$$\det(A+B) \ge \det(A)$$

This would also be enough. Thanks.

• @RodrigodeAzavedo There is a geometric interpretation for this fact: If A and B are positive invertible matrices with $0<A<B$ then $Det A \leq Det B$. We interpret $A,B$ as inner products(Positive symmetric 2-tensores). Then there is a linear map $T$ with $T^*(B)=A)$ this means that T is metric decreasing. But every metric decreasing is a volum decreasing map. Since the n dimensional "Volume" is made by product of 1 dimensional length. Jun 30 '18 at 10:37
• @AliTaghavi Could you please explain what $T$ has got to do with the question? Dec 12 '18 at 21:33

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).

• looks nice, but isn't it somewhat overkill to invoke the Minkowski theorem here? May 19 '11 at 17:05
• 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 $A^{-1/2}BA^{-1/2}$. Jun 28 '14 at 20:02
• I'm going to disagree with @Suvrit and Andreas and say that the Minkowski theorem should be discussed. The reason is that it is a tighter bound that respects the dimensionality, and in particular is saturated when $A=B$ for any $n$. The OP's question is just an awkwardly weakened version of Minkowski. May 22 '15 at 20:10
• In fact, one can show a bit more with @AndreasThom idea: if $A$ is positive definite and $B$ is positive semidefinite and nonzero then $\det(A+B)>\det(A)+\det(B)$ (strict inequality), since $A^{-1/2}BA^{-1/2}$ is then nonzero positive semidefinite, hence with nonnegative eigenvalues and not all 0, and thus $\prod_i(1+\mu_i)\geq 1+\sum_i\mu_i+\prod_i\mu_i>1+\prod_i\mu_i$. Oct 10 '17 at 18:44
• I used $A(1+A^{-1/2}BA^{-1/2}) = A^{1/2}(A+B)A^{-1/2}$. For the second part, you would need to argue that the eigenvalues of $A^{-1}B$ are positive, which brings you back to my argument (I guess). Sep 26 '19 at 6:03

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 $det(A+B)\ge det(A)$.

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)

Update. To expand upon my comment below, to note that the above idea actually with a little bit more care actually yields a proof of the Minkowski determinant inequality, by equivalently establishing log-concavity of the determinant. The key point to observe is \begin{eqnarray} \exp(-x^T((1-\lambda)A+\lambda)x) &=& [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\\\\ \int\exp(-x^T((1-\lambda)A+\lambda)x)dx &=& \int [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\ dx\\\\ &\stackrel{\text{Hölder}}{\le}& \left(\int\exp(-x^TAx)dx \right)^{1-\lambda}\left(\int \exp(-x^TBx)dx \right)^\lambda. \end{eqnarray} Now invoke the Gaussian integral as above to conclude \begin{equation*} \det((1-\lambda)A+\lambda B) \ge \det(A)^{1-\lambda}\det(B)^\lambda, \end{equation*} from which we can easily conclude $\det(A+B)^{1/n} \ge \det(A)^{1/n}+\det(B)^{1/n}$.

• This is very clever. I wonder if such reasoning can also prove the Minkowski determinant theorem mentioned above... Jun 30 '18 at 22:35
• @AlexArvanitakis indeed the same reasoning as above with a tiny bit of care can be used to prove the inequality $\det((1-\lambda)A+\lambda B) \ge \det(A)^{1-\lambda}\det(B)^\lambda$, which is actually equivalent to Minkowski's determinant inequality (easy to verify using suitably scaled versions of the matrices $A$ and $B$) Jul 1 '18 at 22:07

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 ${\rm det}(I+X) \geq (1+d^{1/n})^{n}.$

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.

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 $\mathcal B(B)$.

• It's worth noting that this is secretly the same as Suvrit's answer. 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. May 25 '11 at 7:18
• Fair enough. May 29 '11 at 0:49

Just for those who might want to know, I think the min-max theorem mentioned by @Michael Renardy is Courant-Fischer: Supporse a real symmetric matrix A's eigen values are $$\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n$$. then k-th eigen value: $$\lambda_k = \underset{V_k}{\text{min}} \space max\{x^TAx | x \in V_k, ||x||_2 = 1\}$$ $$V_k$$ is any k-dimentional subspace of $$R^n$$, $$1 \leq k \leq n$$ .