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I want to simplify or find an upper bound for the determinant $|K_1+K_2+I|$ where $I$ is identity matrix, $K_1$ and $K_2$ are positive semi-definite matrices of size $n$ and thus can be written as $K_1=U_1\Lambda_1 U_1^t$, $K_2=U_2\Lambda_2 U_2^t$, in which $U_1$ and $U_2$ are unitary matrices and $\Lambda_1, \Lambda_2$ are diagonal matrices.

Can one say $\det(U_1\Lambda_1 U_1^t +U_2\Lambda_2 U_2^t +I)\le \det(\Lambda_1 +\Lambda_2 +I)$?

Hint: when we have only $K_1$ (or $K_2$) we can write:

$|K_1+I|=|U_1\Lambda_1 U_1^t +I|=|\Lambda_1 U_1^tU_1 +I|=|\Lambda_1 +I|$

where the second equality is due to the identity $\det(AB+I)=\det(BA+I)$.

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    $\begingroup$ Although I'd like to see a counter-example worked out, at first glance this paper seems to say a large bound is needed: "Bounds for the determinant for the sum of Hermitian Matrices" ams.org/journals/proc/1971-030-01/S0002-9939-1971-0286814-1/… Maybe I'm reading the theorem in it incorrectly. $\endgroup$
    – muaddib
    Commented May 27, 2015 at 2:34
  • $\begingroup$ Thank you for this reference. With that I got how to prove the statement. $\endgroup$
    – Bob
    Commented May 27, 2015 at 3:23
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    $\begingroup$ The inequality is obviously incorrect. To correct it, some sort of majorisation or ordered arranegment for the diagonal entries of $\Lambda_1$ and $\Lambda_2$ are needed. $\endgroup$
    – user1551
    Commented May 27, 2015 at 4:15

1 Answer 1

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Unfortunately, the inequality you seek to prove seems not to be established for the general positive semi-definite matrices. I can only prove it for very specific case of $K_1, K_2$ such that eigenvalues of the sum of these matrices are always less than or equal to the corresponding sum of eigenvalues of each of them.

Below is the proof for this special case, if it is of any use for you. For more details see this article.

Recall:

  • The set of positive semidefinite matrices is convex (see section "Further properties", property 12), i.e. $K_1, K_2$ are positive semidefinite $\implies \alpha K_1 + (1-\alpha) K_2$ is also positive semidefinite for all $\alpha \in (0,1)$. By setting $\alpha = \frac{1}{2}$, we establish that the sum of two positive semidefinite matrices $\left(K_1 + K_2\right) : = L $ is also positive semidefinite.

  • For two positive semidefinite matrices $A,B$ the following inequality folds: $\det(A+B)\ge \det(A) + \det(B)$. This is due to Minkowski's theorem

I assume that by $\det(U_1\Lambda_1 U_1^t +U_2\Lambda_2 U_2^t +I)\le \det(\Lambda_1 +\Lambda_2 +I)$ you mean
$$ \left| K_1 + K_2 + I \right| \le \left| \Lambda_1 + \Lambda_2 + I \right| $$ where $K_1=U_1\Lambda_1 U_1^t, K_2=U_2\Lambda_2 U_2^t$.

Since $ L:=K_1 + K_2$ is positive semidefinite, it can be diagonalized as $L = P D P^T$. Then $$ \left| K_1 + K_2 +I \right| = \left| L +I \right| = \left| D + I \right| = \prod_{i=1}^n \big( \lambda_i\left( \right) + 1 \big), $$

where $\lambda_1(Q), \dots, \lambda_n(Q)$ are the eigenvalues of a matrix $Q$. On the other hand,

$$ \left| \Lambda_1 + \Lambda_2 + I \right| = \prod_{i=1}^n \big( \lambda_i\left( K_1\right) + \lambda_i\left( K_2\right) + 1\big) $$

If for all $ i = 1, \dots,n$ we had the following inequality $$ \lambda_i(L) =\lambda_i(K_1 + K_2) \le \lambda_i(K_1) +\lambda_i(K_2), $$ it would imply $$ \left| K_1 + K_2 +I \right| = \prod_{i=1}^n \big( \lambda_i\left( L\right) + 1 \big) \le \prod_{i=1}^n \big( \lambda_i\left( K_1\right) + \lambda_i\left( K_2\right) + 1\big) = \left| \Lambda_1 + \Lambda_2 + I \right| $$

If the eigenvalues inequality does not hold, then the result would not be valid.

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