How does one prove the determinant inequality $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$?

Question

_{Reposted on MathOverflow}

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Let $\,A,B,C\in M_{n}(\mathbb C)\,$ be Hermitian and positive definite matrices such that $A+B+C=I_{n}$, where $I_{n}$ is the identity matrix. Show that $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n \det \left(A^2+B^2+C^2\right)$$

This problem is a test question from China (xixi). It is said one can use the equation

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

but I can't use this to prove it. Can you help me?

Answer 1

Here is a partial and positive result, valid around the "triple point" $A=B=C= \frac13\mathbb 1$.

Let $A,B,C\in M_n(\mathbb C)$ be Hermitian satisfying $A+B+C=\mathbb 1$, and additionally assume that $$\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\, \|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}$$ in the spectral or operator norm. (In particular, $A,B,C$ are positive-definite.)
Then we have $$6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}$$

Proof: Let $A_0=A-\frac13\mathbb 1$ a.s.o., then $A_0+B_0+C_0=0$, or $\,\sum_\text{cyc}A_0 =0\,$ in notational short form. Consider the

• Sum of squares $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big) \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1$$
• Sum of cubes $$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 \:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 + 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\ \;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1$$ to obtain $$6\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1 \;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 0$$ where positivity is due to each summand being a product of commuting positive-semidefinite matrices. $\quad\blacktriangle$

Two years later observation:
In order to conclude $(2)$ the additional assumptions $(1)$ may be weakened a fair way off to $$\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}$$ or equivalently, assuming the smallest eigenvalue of each matrix $A,B,C\,$ to be at least $\tfrac16$.

Proof: Consider the very last summand in the preceding proof. Revert notation from $A_0$ to $A$ and use the same argument, this time based on $(3)$, to obtain $$\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle$$

Answer 2

Not a full proof, but a number of thoughts too long for a comment. This post aims at finding alternative (yet harder) criteria for proving the conjecture. Please discuss.

As in previous comments, let's denote $ X=6(A^3+B^3+C^3)+I $ and $ Y=5(A^2+B^2+C^2) $ and $D = X-Y$.

The question is to show $\det(X) \ge \det(Y)$ or $1 \ge \det(X^{-1}Y)$. Write $Q = X^{-1}Y$, then $Q$ is positive definite, since $X$ and $Y$ are positive definite. Now it is known for a positive definite matrix $Q$ (see e.g. here) that the trace bound is given by $$ \bigg(\frac{\text{Tr}(Q)}{n}\bigg)^n \geq \det(Q) $$

So a second (harder) criterion for the conjecture is $n \ge \text{Tr}(X^{-1}Y)$ or $ \text{Tr}(X^{-1}D) \geq 0$. I wouldn't see how I can compute this trace or find bounds, can someone?

Let's call $d_i$ the eigenvalues of $D$, likewise for $X$ and $Y$. While $x_i > 0$, this doesn't necessarily hold for $d_i$ since we know from comments that $D$ is not necessarily positive definite. So (if $X$ and $D$ could be simultaneously diagonalized) $ \text{Tr}(X^{-1}D) = \sum_i \frac{d_i}{x_i} = r \sum_i {d_i}$ where there exists an $r$ by the mean value theorem. Where $r$ is not guaranteed to be positive, it is likely that $r$ will be positive, since $r$ will only become negative if there are (many, very) negative $d_i$ with small associated $x_i$. Can positivity of $r$ be shown? If we can establish that a positive $r$ can be found, a third criterion is $ \text{Tr}(D) \geq 0$.

Now with this third criterion, we can use that the trace is additive and that the trace of commutors vanishes, i.e. $\text{Tr} (AB -BA) = 0$. Using this argument, it becomes unharmful when matrices do not commute, since under the trace their order can be changed. This restores previous solutions where the conjecture was reduced to the valid Schur's inequality (as noted by a previous commenter), which proves the conjecture.

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A word on how hard the criteria are, indicatively in terms of eigenvalues:

(hardest) positive definiteness: $d_i >0$ $\forall i$ or equivalently, $\frac{y_i}{x_i} <1$ $\forall i$

(second- relies on positive $r$) $ \text{Tr}(D) \geq 0$: $\sum_i d_i \geq 0$

(third) $n \ge \text{Tr}(X^{-1}Y)$: $\sum_i \frac{y_i}{x_i} \leq n$

(fourth - least hard) $\det(X) \ge \det(Y)$: $\prod_i \frac{y_i}{x_i} \leq 1$

Solutions may also be found by using criteria which interlace between those four.

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A word on simulations and non-positive-definiteness:

I checked the above criteria for the non-positive definite example given by @user1551 in the comments above, and the second, third and fourth criteria hold.

Note that equality $\det(X) = \det(Y)$ occurs for (a) symmetry point: $A=B=C=\frac13 I$ and for (b) border point: $A=B=\frac12 I$ and $C=0$ (and permutations). I checked the "vicinity" of these equality points by computer simulations for real matrices with $n=2$ where I extensively added small matrices with any parameter choices to $A$ and $B$ (and let $C = I - A-B$), making sure that $A,B$ and $C$ are positive definite. It shows that for the vicinity of the symmetry point, the second, third and fourth criteria above hold, while there occur frequent non-positive-definite examples. For the vicinity of the border point all four criteria hold.

Answer 3

Faulty Proof

In the comments we have a proof over positive real numbers $a,b,c \geq 0$ that the inequality holds

$$ 6 \big( a^3 + b^3 + c^3 \big) + 1 \geq 5(a^2 + b^2 + c^2) $$

We can lift this to diagonal matrices, but how to we lift this to Hermitian positive definite matrices? Hermitian already gives you $\lambda \in \mathbb{R}$. I think that is missing is that

$$ X = \Big[ 6 \big( A^3 + B^3 + C^3 \big) + I_n \Big]- \Big[ 5(A^2 + B^2 + C^2) \Big] \geq 0 $$

is also positive definite. In particular one shows $\mathbf{v}^T X \mathbf{v} \geq 0$ for all unit vectors $|| \mathbf{v} || = 1$

$$ \lambda_{min} = \inf_{||v|| =1} \mathbf{v}^T X \mathbf{v} $$

This is the variational characterization of the smallest eigenvalue. Through elaborate checks and balances one can obtain * all * eigenvlues. [1] (Not needed here, but one time I needed to compute $\lambda_2$)

There are three positive definite matrices. $X = X_1 - X_2$ where

• $X_1 = 6 \big( A^3 + B^3 + C^3 \big) + 1 \geq 0$
• $X_2 = 5 \big( A^2 + B^2 + C^2 \big) \quad \;\, \geq 0$

all of this follows from the variational inequality. Last we can take the determinant of both sides:

$$ \det \bigg[ 6 \big( A^3 + B^3 + C^3 \big) + 1\bigg] \geq \det (X_1 - X_2) + \det X_2 \geq 5^n \det \bigg[ \big( A^2 + B^2 + C^2 \big) \bigg] $$

There are a few discussions of inequalities of this type on MathOverflow

• https://mathoverflow.net/q/65424
• https://mathoverflow.net/q/182181

$\det (X+Y) \geq \det X + \det Y $ follows from Minkowski inequality.

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Other Ideas

I have thought of several faulty proofs over night but here is one observation:

• all eigenvalues $0 \leq \lambda_A, \lambda_B, \lambda_C\leq 1 $

I think this is right. Certainly they are all $\geq 0$. And then we can use variational inequality

$$ 0 \leq \lambda_A + \lambda_B + \lambda_C \leq v^T ( A+B+C) v = v^T v = 1$$

There is even easier argument if we take the trace and there is equality.

$$ \text{Tr}(A+B+C) = \sum \lambda_A + \sum \lambda_B + \sum \lambda_C = n $$

This is very restrictive (as Hans said) we get that $|\lambda_i - \lambda_j| \leq 1$ always and a fair bit more. In particular, the max and the min are separated by less than 1.

Remember: in many cases faulty proofs can be turned into correct ones! Thanks.

Answer 4

This is not a complete answer but a summary of progress I have made. I suspect the primary author was aware of results related to those I am about to present and decided to include only a special case, perhaps in the expectation that resolving that case would help advance the full conjecture. I now state a conjecture and provide further details and intuition.

Let the power-sum symmetric polynomial of degree $k$ in $n$ variables $x_1,\ldots,x_n$ be denoted by $P_k$, defined by

$$ P_k(x_1,\ldots,x_n)=\sum_{i=1}^n x_i^k. $$

Below I state a conjecture which I believe may hold under additional assumptions. I remain open to the possibility that the conjecture, as stated, may be false; however, I expect that suitable extra hypotheses will make it valid.

Conjecture 1. Let $(A_i)_{i=1}^{m}$ be a collection of $n \times n$ positive definite Hermitian matrices such that $\sum_{i=1}^{m} A_i = I_n$. For each matrix $A_i$, let $\lambda(A_i) = (\lambda_1(A_i), \ldots, \lambda_n(A_i))$ denote the corresponding eigenvalues of the matrix, ordered decreasingly. For $ \alpha_i,\beta_i \in \mathbb{R} $,let \begin{align*} f(A_1,\ldots,A_m) = \sum_{i=1}^{m} \alpha_i P_i(A_1,\ldots,A_m) \\ g(A_1,\ldots,A_m) = \sum_{i=1}^{m} \beta_i P_i(A_1,\ldots,A_m) \end{align*} be symmetric $m$ variable matrix functions satisfying the scalar inequality, meaning that for any $x_1,\ldots,x_m \in \mathbb{R}^{+}$ such $\sum_{i=1}^{m} x_i = 1$ we have $$ 0 \le g(x_1,\ldots,x_m) \le f(x_1,\ldots,x_m). $$ Then $$ \lambda\left( g(A_1, \ldots, A_m) \right) \prec_{w} \lambda\left( f(A_1, \ldots, A_m) \right). $$ By $\prec_{w}$ we mean weakly majorization.

If the conjecture above is valid, then a suitable modification of the following result yields a stronger statement.

Theorem 1. Let $n\ge 1$, let $d\ge 0$ be even, and let $x_1,\dots,x_n$ be real numbers. If $(x_1,\dots,x_n)$ majorizes $(y_1,\dots,y_n)$, then

$$ h_d(x_1,\dots,x_n)\ge h_d(y_1,\dots,y_n), $$

with equality if and only if $(y_1,\dots,y_n)$ is a permutation of $(x_1,\dots,x_n)$.

Moreover, if one restricts the variables to be positive and replaces majorization by weak majorization, the same argument leads to stronger matrix inequalities (in particular, one obtains corresponding inequalities for determinants and other spectral quantities).

To support these claims I have tested the conjecture numerically for the original question and have not found counterexamples so far. Another example I examined is the following

$$ f(A,B,C)=\frac{116}{81}I_n + 64\big(A^5+B^5+C^5\big),\qquad g(A,B,C)=20\big(A^3+B^3+C^3\big), $$

under the usual assumptions. For these choices, $f-g$ need not be positive semidefinite in general; an explicit $2\times2$ counterexample is given by

$$ A=\begin{bmatrix}0.363 & 0\\[4pt]0 & 0.767\end{bmatrix},\quad B=\begin{bmatrix}0.306 & -0.13534244755789848\\[4pt]-0.13534244755789848 & 0.075\end{bmatrix},\quad C=\begin{bmatrix}0.331 & 0.13534244755789848\\[4pt]0.13534244755789848 & 0.158\end{bmatrix}, $$

for which $f-g$ is not necessarily positive semidefinite.

One natural tool to attempt here is the following variational characterization of partial sums of eigenvalues.

Theorem 2 (Ky-Fan). Let $A$ be an $n\times n$ Hermitian matrix with eigenvalues $\lambda_1(A)\ge\lambda_2(A)\ge\cdots\ge\lambda_n(A)$. Then, for each $1\le k\le n$,

$$ \max_{Y\in\mathbb{C}^{n\times k},\;Y^*Y=I_k}\operatorname{Tr}(Y^* A Y)=\sum_{i=1}^k \lambda_i(A), $$

where the maximum is taken over all $n\times k$ matrices $Y$ whose columns are orthonormal.

The advantage of this theorem is that, for a given choice of $Y$, the compressed matrices $(Y^*A_iY)_{i=1}^m$ again sum to $I_k$, so one can reduce dimension and attempt an inductive or variational approach. I was, however, unable to make substantial progress along this line. One important point to note is that for any minimizer or maximizer $Y$, its columns span an invariant subspace of $A$ associated with the first $k$ eigenvalues, check On Generalizing Trace Minimization.

For the final part, I will show that the following theorem is correct. It represents a very special case of the material discussed above.

Theorem 3. By the stated assuptions one have $$ \text{Tr}(5(A^2+B^2+C^2)) \le \text{Tr}(I_n+6(A^3+B^3+C^3)) $$

Proof . Since $ \text{Tr}(A)+\text{Tr}(B)+\text{Tr}(C)=\text{Tr}(I_n) $, a matrix exists such that it's trace is strictly less than $\text{Tr}( I_n/3 )$. Otherwise $$ \text{Tr}(I_n) = \text{Tr}(\frac{I_n}{3})+\text{Tr}(\frac{I_n}{3})+\text{Tr}(\frac{I_n}{3}) < \text{Tr}(A)+\text{Tr}(B)+\text{Tr}(C) . $$ Which is not possible. Hence, without loss of generality we can assume that $\text{Tr}(C) \le \text{Tr}(\frac{I_n}{3})$. Additionally consider the Eigendecomposition of $C$ as follow $$ C=U^{*}C^{'}U,U^{*}U=I_n. $$ Where $C^{'}$ is a diagoanl matrix. Set $$ X_n = I_n+6(A^3+B^3+C^3), Y_n = 5(A^2+B^2+C^2) . $$ It's straightforward to verify that $$ \text{Tr}(X_n)=\text{Tr}(U^{*}X_nU),\text{Tr}(Y_n)=\text{Tr}(U^{*}Y_nU) . $$ We additionally establish $$ U^{*}AU+U^{*}BU+C{'} = U^{*}(A+B+C)U = U^{*}U = I_n . $$ As a result, it's enough to prove that \begin{align*} 0 \le \text{Tr}(U^{*}X_nU) - \text{Tr}(U^{*}Y_nU) . \end{align*} Let us define \begin{align*} A^{'} &= \text{diag}(U^{*}AU) \\ B^{'} &= \text{diag}(U^{*}BU) \\ K &= U^{*}AU-A^{'} . \end{align*} Clearly $$ (A^{'}+K) + (B^{'}-K) + C^{'} = I_n . $$ From the preceding results, we obtain \begin{align*} & \text{Tr}(U^{*}X_nU)-\text{Tr}(U^{*}Y_nU) \\ & = \text{Tr}(I_n+6((U^{*}AU)^3+(U^{*}BU)^3+C'^3)-5((U^{*}AU)^2+(U^{*}BU)^2+C'^2)) \\ & = \text{Tr}(I_n+6((A^{'}+K)^3+(B^{'}-K)^3+C'^3) -5((A^{'}+K)^2+(B^{'}-K)^2+C'^2)) \\ & = \text{Tr}(I_n+6(A'^3+B'^3+C'^3)-5(A'^2+B'^2+C'^2))+6\text{Tr}(K^3-K^3)+6\text{Tr}(3(A'^2-B'^2)K) \\ & + 6\text{Tr}(3(A'+B')K^2)-10\text{Tr}(K^2)-10\text{Tr}((A'-B')K) . \end{align*} Since $K$ is a Hermitian matrix and it's diagonal elements are zero and matrices $A^{'}$ and $B^{'}$ are diagonal, we can deduce $$ 6\text{Tr}(3(A'^2-B'^2)K)=10\text{Tr}((A'-B')K)=6\text{Tr}(K^3-K^3)=0 . $$ Consequently \begin{align*} & \text{Tr}(U^{*}X_nU)-\text{Tr}(U^{*}Y_nU) \\ & =\text{Tr}(I_n+6(A'^3+B'^3+C'^3)-5(A'^2+B'^2+C'^2))+\text{Tr}((18(A'+B')-10I_n)K^2). \end{align*} The next step is to show that $$ 0 \le \text{Tr}((18(A'+B')-10I_n)K^2) . $$ To prove this, please note that $A^{'}+B^{'} = I_n- C^{'}$ and $\text{Tr}(C^{'}) \le \text{Tr}(I_n/3)$. Resulting in \begin{align*} \text{Tr}((18(A'+B')-10I_n)K^2) & \ge \\ \text{Tr}((18(I_n-C')-10I_n)K^2) & \ge\\ \text{Tr}((18\times \frac{2}{3}I_n-10I_n)K^2) & \ge 0 . \end{align*} It's left to demonstrate that $$ \text{Tr}(I_n+6(A'^3+B'^3+C'^3)-5(A'^2+B'^2+C'^2)) \ge 0 . $$ Which is equivalent to the scalar case, this completes the proof.