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To all those who are eagerly awaiting a new question, all those who love math, I give this challenge and I hope for you good moments of reflection.

Let $A=(a_{ij})_n$ a real nonnegative symmetric matrix with eigenvalues $\lambda_1\geq\cdots\geq \lambda_n\geq 0$.

How to prove that, for all $k\in\{1,\ldots,n\}$, we have $$\prod_{j=1}^ka_{jj}\leq\left(\frac{1}{k}\sum_{j=1}^k\lambda_j\right)^k.$$

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With "real positive" do you mean $a_{ij}>0$ for all $i,j$? –  Matemáticos Chibchas Feb 27 '13 at 16:25
That means $\langle Ax,x\rangle\ge 0$@MatemáticosChibchas –  Yimin Feb 27 '13 at 16:26
Try to diagonalize the matrix from right bottom to left top. –  Yimin Feb 27 '13 at 16:35
for $A$ is positive matrix, then $\lambda_i > 0$ ?. –  Laura Feb 27 '13 at 16:38
@Tai, you're right, I mean $A$ nonnegative. –  Sami Ben Romdhane Feb 27 '13 at 16:50

3 Answers 3

Apply G.M. $\le$ A.M. to $a_{jj}, j = 1,\ldots,k$, we have:

$$\prod_{j=1}^k a_{jj} \le \left( \frac{1}{k} \sum_{j=1}^k a_{jj} \right)^{k}$$

Since $A$ is a real non-negative symmetric matrix with non-negative eigenvalues $\lambda_1 \ge \lambda_2 \ge \cdots \lambda_n \ge 0$, we can find an orthogonal matrix $\Omega$ and diagonal matrix $\Lambda$ such that $A = \Omega^{T} \Lambda \Omega$ and $\Lambda_{ii} = \lambda_i$ for $i = 1,\ldots,n$. In terms of coefficients of $\Omega$, we have:

$$\sum_{j=1}^k a_{jj} = \sum_{j=1}^k \sum_{i=1}^n \lambda_i |\Omega_{ij}|^2 =\sum_{i=1}^n\lambda_i \beta_i \,\,\,\text{ where }\,\,\,\beta_i = \sum_{j=1}^k |\Omega_{ij}|^2 $$

Using properties of orthogonal matrices, it is not hard to see $0 \le \beta_i \le 1$ and $\sum_{i=1}^n \beta_i = k$.

Let $x_0 = 0$ and $x_i = \sum_{s=1}^i \beta_s$ for $i = 1,\ldots,n$. Define two functions $\beta(t)$, $\gamma(t)$ on $[0,k)$ by:

$$\begin{align} \beta(t) &= \lambda_i \,\,\,\text{ for }\,\,\, t \in [x_{i-1}, x_i), & i = 1,\ldots, n\\ \gamma(t) &= \lambda_i \,\,\,\text{ for }\,\,\, t \in [i-1, i), & i = 1,\ldots, k \end{align}$$

Since $\lambda_i$ is non-increasing, it is easy to see $\beta(t) \le \gamma(t)$ on $[0,k)$. From this, we can deduce:

$$\sum_{i=1}^{n} \lambda_i\beta_i = \int_{0}^{k} \beta(t) dt \le \int_{0}^{k} \gamma(t) dt = \sum_{i=1}^k \lambda_i$$

As a result, we can conclude: $$\prod_{j=1}^k a_{jj} \le \left( \frac{1}{k} \sum_{j=1}^k \lambda_j \right)^{k}$$

EDIT Background information.

The statement $\sum_{j=1}^k a_{jj} \le \sum_{j=1}^k \lambda_j$ is actually a corollary of a well known theorem first proved by Schur. Namely,

Let $A$ be an $n\times n$ Hermitian matrix. Let $\text{diag}(A)$ denote the vector whose coordinates are the diagonal entries of $A$ and $\lambda(A)$ a vector whose coordinates are the eigenvalues of $A$ arranged in any order, then $\text{diag}(A)$ is majorized by $\lambda(A)$.

What this means is when we sort the components of $\text{diag}(A)$ and $\lambda(A)$ into two n-tuples of decreasing order:
$$a^{\downarrow}_1 \ge a^{\downarrow}_2 \ge \cdots \ge a^{\downarrow}_n \,\,\,\text{ and }\,\,\, \lambda^{\downarrow}_1 \ge \lambda^{\downarrow}_2 \ge \cdots \ge \lambda^{\downarrow}_n$$ we have $a^{\downarrow}_1 \le \lambda^{\downarrow}_1$, $\,\,\,a^{\downarrow}_1 + a^{\downarrow}_2 \le \lambda^{\downarrow}_1 + \lambda^{\downarrow}_2$ and in general, $$\sum_{i=1}^k a^{\downarrow}_i \le \sum_{i=1}^k \lambda^{\downarrow}_i \,\,\,\text{ for } 1 \le k < n\,\,\,\text{ and }\,\,\, \sum_{i=1}^n a^{\downarrow}_i = \sum_{i=1}^n \lambda^{\downarrow}_i$$

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Is your definition of $\beta(t)$ correct? Instead of $x_i$ should it be simply $i$? –  robjohn Jun 4 '13 at 2:08
@robjohn, the definition of $\beta(t)$ is correct. Otherwise, you can't reproduce the sum $\sum_{i=1}^{n} \lambda_i\beta_i$. –  achille hui Jun 4 '13 at 2:43
What are the $x_i$ then? They are never mentioned elsewhere. –  robjohn Jun 4 '13 at 4:42
$x_i = \sum_{s=1}^{i} \beta_{s}$, they are introduced just to prove the inequality $$\sum_{i=1}^{n} \lambda_i \beta_i \le \sum_{i=1}^{k} \lambda_i$$ If you don't like this argument, an alternate argument goes like this. First consider the special case when all $\beta_i \in \mathbb{Q}$, then the inequality is true because it can be recasted as some form of rearrangement inequality. One then prove the general case by treating it as some sort of limit of rational approximations of $(\beta_i)_{i=1,\ldots,n}$. –  achille hui Jun 4 '13 at 5:01
Ah, sorry, I missed the definition of the $x_i$. I see it now. –  robjohn Jun 4 '13 at 5:03

These inequalities hold more generally for $A\in M_n(\mathbb{C})$ hermitian semidefinite positive. My favourite argument uses the most convenient min-max theorem, which says that when the eigenvalues of $A$ are in nonincreasing order $\lambda_1\geq \lambda_2\geq\ldots \geq \lambda_n\geq 0$, we have $$ \lambda_j=\max_{\dim F=j}\;\;\min_{x\in F, \|x\|=1} (Ax,x) $$ where the max is taken over all dimension $j$ subspaces $F$ of $\mathbb{C}^n$.

Denoting $\{e_1,\ldots,e_n\}$ the canonical basis of $\mathbb{C}^n$, we see that $a_{jj}=(e_j,Ae_j)\geq 0$ for every $j$. By AM-GM applied to $a_{11},\ldots,a_{kk}$, we see that it suffices to prove that $$ \sum_{j=1}^k a_{jj}\leq \sum_{j=1}^k\lambda_j \qquad\forall 1\leq k\leq n. $$ In the case $k=n$, we easily see that this is an equality as both numbers are equal to the trace of $A$. It is more delicate to handle the case $k\leq n-1$.

Now we denote $A_k$ the upper-left square block of $A$ of size $k$, i.e. $A_k=(a_{ij})_{1\leq i,j\leq k}$. Note that it is a hermitian positive semidefinite matrix in $M_k(\mathbb{C})$ as $(A_kx,x)=(Ax,x)$ for every $x\in \mathbb{C}^k$ identified with the subspace of $\mathbb{C}^n$ spanned by $e_1,\ldots,e_k$. Writing $\mu_1\geq\ldots\geq \mu_k\geq 0$ the eigenvalues of $A_k$, the min-max theorem applied to $A_k$ yields $$ \mu_j=\max_{F\subseteq \mathbb{C}^k\dim F=j}\;\;\min_{x\in F, \|x\|=1} (A_kx,x)=\max_{F\subseteq \mathbb{C}^k\dim F=j}\;\;\min_{x\in F, \|x\|=1} (Ax,x). $$ Since the maximum of the rhs runs over a subset of the set of all dimension $j$ subspaces of $\mathbb{C}^n$, it is not greater than the maximum over the latter. Therefore $$ \forall 1\leq j\leq k\qquad \mu_j\leq \lambda_j\quad\Rightarrow\quad \sum_{j=1}^ka_{jj}=\mbox{tr}A_k=\sum_{j=1}^k\mu_j\leq \sum_{j=1}^k\lambda_j. $$

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This is close to how I was working on this, so perhaps I will leave my answer deleted unless I come up with something significantly different. –  robjohn Jun 4 '13 at 15:30
Fair enough. I was thinking of the full inequality, not the intermediate equality. –  robjohn Jun 4 '13 at 17:27

Hint: Suppose orthogonal matrix $P_k$ can diagonalize the matrix $A_{k}$, where $A_k$ means the matrix made by the first $k$ cols and $k$ rows of $A$.

As $P_kA_kP_k' = \Phi_k $ as a diagonal matrix, with diagonal as $\left[\mu_1,\mu_2,\cdots,\mu_k\right]$, arranged in decreasing order.

write $A$ as

$\left(\begin{array}[cc] AA_k&U\\ U'&R \end{array}\right)$

where $U$ as $k\times (n-k)$ matrix. $R$ as symmetric $(n-k)\times(n-k)$ matrix.

$\left(\begin{array}[cc] AP_k&0\\ 0&I_{n-k} \end{array}\right)$$\left(\begin{array}[cc] AA_k&U\\ U'&R \end{array}\right)$ $\left(\begin{array}[cc] AP_k'&0\\ 0&I_{n-k} \end{array}\right)$ = $\left(\begin{array}[cc] A\Phi_k& P_kU\\ U'P_k'&R \end{array}\right)$

And $tr(\Phi_k) = tr(A_k) = \sum_{i=1}^k a_{ii}$, take $\Lambda = \left[\lambda_1,\cdots,\lambda_n\right]$

If we can prove $\sum_{i=1}^k\lambda_i\ge tr(\Phi_k)$, then we can use AM-GM inequality to prove it.

take $x=(x_1,\cdots,x_k,0)$, $0$ is a vector in dimension $n-k$.

Consider $z =x\left(\begin{array}[cc] AP_k'&0\\ 0&I_{n-k} \end{array}\right) P'_n $,

then $\sum_{i=1}^n\lambda_i z_i^2 =\sum_{j=1}^k\mu_j x_j^2 $,

we take $x = \left[0,0,\cdots,1,0,0\cdots,0\right]$ for first $k$ basis.

when $x = e^m$, as $m$th basis, we got $z^m = (z_1^m,\cdots,z_n^m)$.

And it is easy to prove $z^l$,$z^m$ are orthogonal when $m\neq l$.

Therefore $\sum_{j=1}^k(z_i^j)^2\le 1$. [Note arrange all the $z^l$ as a matrix, and see this by looking at cols]


$\sum_{i=1}^n \lambda_i (z_i^m)^2 = \mu_m$,


$\sum_{j=1}^k \mu_j = \sum_{i=1}^n\lambda_i(\sum_{j=1}^k(z_i^j)^2)\le \sum_{i=1}^k \lambda_i$.


$\sum_{i=1}^n\sum_{j=1}^k(z_i^j)^2 = k$.

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