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$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\tr}{\operatorname{tr}}$

Let $d \in \mathbb{N}, d \ge 2$. Let $D \in M_d$ be a non-zero diagonal matrix.

Consider the equation $\Sigma^2=(\det \Sigma)^{\frac{2}{d}}\id+D$ where $\Sigma$ is a diagonal real $d \times d$ matrix with non-negative values.

I found two necessary conditions for the existence of a solution:

(1) $\tr(D) > 0$.

(2) Not all the diagonal elements (of $D$) are positive.

(Proofs are at the end).

Question: Suppose $\tr(D) > 0$, and that not all the elements of $D$ are positive. Is there a solution? How many solutions are there?

Can the solutions be expressed "nicely" in terms of $D$?

For the case $d=2$, I showed below that indeed these conditions are sufficient, and when they hold there is a unique solution:

Denoting $D=\begin{pmatrix} \lambda_1 & 0 \\\ 0 & \lambda_2 \end{pmatrix}$, the solution is given by

$$\Sigma=\begin{pmatrix} \frac{|\lambda_1|}{\sqrt{\tr(D)}} & 0 \\\ 0 & \frac{|\lambda_2|}{\sqrt{\tr(D)}} \end{pmatrix} \tag{1}=\frac{|D|}{\sqrt{\tr(D)}}.$$

Any ideas how to solve the general case?

(For the interested, I got to this equation after studying the modified conformality equation $ A^TA=|\det A|^{\frac{2}{d}}\id+B$ and using its orthogonal invariance).

Proofs of the necessary conditions:

(1): $\tr(D) > 0$.

Indeed, denoting $\Sigma=\operatorname{diag}(\sigma_1,...,\sigma_d)$, $\sigma_i^2=\alpha_i$ we take the trace: $$ \tr(\Sigma^2)=\sum_{i=1}^d \alpha_i = d(\Pi_{i=1}^d \alpha_i)^{\frac{1}{d}}+\tr(D),$$ hence $$ \frac{\tr(D)}{d} =\frac{\sum_{i=1}^d \alpha_i}{d}-(\Pi_{i=1}^d \alpha_i)^{\frac{1}{d}} \ge 0,$$ and equality holds if and only if $\alpha_1=\alpha_2=\dots=\alpha_d$ (This is the AM–GM inequality).

In the equality case, we get $\alpha_i=\alpha,\sigma_i=\sigma=\sqrt{\alpha}$, so $\Sigma =\sigma \id$ is conformal, so $D=0$ which contradicts our assumption.

(2): Not all the diagonal elements are positive.

Denote $D=\operatorname{diag}(\lambda_1,...,\lambda_d)$. Then the equation is $$\sigma_i^2-(\Pi_{i=1}^d \sigma_i)^\frac{2}{d}=\lambda_i,i=1,\dots,d$$

If all the $\lambda_i$ are positive, then each $\sigma_i$ is greater than their geometric mean, which is impossible.


General observation: Denoting $\Sigma=\operatorname{diag}(\sigma_1,...,\sigma_d),D=\operatorname{diag}(\lambda_1,...,\lambda_d)$ the equation reduces to: $$\sigma_i^2-(\Pi_{i=1}^d \sigma_i)^\frac{2}{d}=\lambda_i, i=1,...,d$$

In particular $\sigma_1^2-\sigma_j^2=\lambda_1-\lambda_j$. Since $\sigma_i \ge 0$, $\sigma_1$ determines all the other $\sigma_j$.


Analysis of the $2D$ case:

Recall that $D=\begin{pmatrix} \lambda_1 & 0 \\\ 0 & \lambda_2 \end{pmatrix}$, and that we assumw $\tr(D)=\lambda_1+\lambda_2 >0$:

$\sigma_i^2- \sigma_1\sigma_2=\lambda_i,$ so $ \sigma_1^2-\sigma_2^2=\lambda_1-\lambda_2,(\sigma_1-\sigma_2)^2=\lambda_1+\lambda_2$.

Since we assumed $D \neq 0$, $\sigma_1 \neq \sigma_2 $ so we divide an obtain $$ \frac{\sigma_1+\sigma_2}{\sigma_1-\sigma_2}=\frac{\lambda_1-\lambda_2}{\lambda_1+\lambda_2}:=\lambda,$$ hence $\sigma_1+\sigma_2=\lambda\sigma_1-\lambda\sigma_2 \Rightarrow \sigma_1(1-\lambda)=-\sigma_2(\lambda+1)$.

Now we divide into cases:

(1) $\lambda_1=0$: This forces $\lambda_2 > 0$. Then $\sigma_1(\sigma_1-\sigma_2)=0 \Rightarrow \sigma_1=0 \Rightarrow \sigma_2^2=\lambda_2$.

In this case $\sigma_2=\sqrt{\lambda_2}$, i.e $\Sigma = \sqrt{D}$.

(2) $\lambda_1 \neq 0$: Then $1+\lambda \neq 0$, so $ \sigma_2=\sigma_1\frac{\lambda-1}{\lambda+1}=-\frac{\lambda_2}{\lambda_1}\sigma_1$.

From the equation $\sigma_1^2- \sigma_1\sigma_2=\sigma_1^2(1+\frac{\lambda_2}{\lambda_1} ) =\sigma_1^2(\frac{\lambda_1+\lambda_2}{\lambda_1} )=\lambda_1$, so

$$\sigma_1^2 = \frac{\lambda_1^2}{\lambda_1+\lambda_2}=\frac{\lambda_1^2}{\tr(D)}.$$ Finally, $$ \sigma_1 = \frac{|\lambda_1|}{\sqrt{\tr(D)}},\sigma_2 = -\frac{\lambda_2}{\sqrt{\lambda_1+\lambda_2}}=\frac{|\lambda_2|}{\sqrt{\tr(D)}}. \tag{2}$$

We observe that the solution in case $(1)$ is in fact in the form of $(2)$.

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This is not a question about linear algebra. Let $D=diag(d_i),\Sigma=diag(\sigma_i),m=(\Pi_i{\sigma_i}^2)^{1/d}$. Necessarily $m\geq \sup_i(-d_i)$ and $\sigma_i=\sqrt{m+d_i}$.

Case 1. $m=0$. Then, for every $i$,$d_i\geq 0$, there is $j,k$ s.t. $d_j=0,d_k\not= 0$ and, for every $i$, $\sigma_i=\sqrt{d_i}$.

Case 2. $m>0$. Then, let $K=\{j|d_j\not= 0\}$. For every $j\in K$, $1+d_j/m>0$ and $\Pi_{j\in K}(1+d_j/m)=1$. We consider the continuous function $$f:x\in[\sup_i(-d_i),+\infty[\rightarrow \Pi_{j\in K}(1+d_j/x).$$ We look for $m$ s.t. $f(m)=1$. When $x\rightarrow +\infty$, $f(x)=1+trace(D)/x+o(1/x)$ and, since $trace(D)>0$, $\lim_{x\rightarrow +\infty}=1+$.

i) For every $i$, $d_i\geq 0$ and there is $j$ s.t. $d_j=0$. Then ${\sigma_j}^2=m$ and for every $i$, ${\sigma_i}^2\geq m$. Consequently, the $(\sigma_i)$'s are equal, that is contradictory (cf. the OP's post).

ii) There is $i$ s.t. $d_i<0$; then, $\sup_i(-d_i)>0$; thus $f(\sup_i(-d_i))=0$ and we are done.

EDIT 1. (To Asaf Shachar). I just read your last post. Unfortunately, $f$ is not monotone because it is decreasing in a neighborhood of $+\infty$ (its derivative is $\approx -trace(D)/x^2$). Then, I proved only that the equation $f(x)=1$ has at least one solution.

EDIT 2. To @ Asaf Shachar. Now I prove (with the help of a colleague) that there is exactly one solution; it suffices to show that the function $$f'/f:x\in ]\sup_i(-d_i),+\infty[\rightarrow \sum_{j\in K}(\dfrac{1}{x+d_j}-\dfrac{1}{x})$$ has exactly one zero.

Proof. We assume that $d_1<\cdots<d_p<0<d_{p+1}<\cdots <d_k,\sum_{i\leq k}\alpha_id_i>0$, where $\alpha_i$ is the "multiplicity" of $d_i$. We consider the function

$$g:x\notin \{-d_j|j\leq k\}\rightarrow \sum_{j\leq k }\dfrac{\alpha_jd_j}{x(x+d_j)}.$$ Note that $g(x)=\dfrac{P_{k-1}(x)}{x\Pi_i(x+d_i)}$, where $degree(P_{k-1})=k-1$, and $g(x)=0$ iff $P_{k-1}(x)=0$, that is, $g$ has at most $k-1$ roots. Considering the ($\lim_{x\rightarrow \pm -d_i}g(x)$)'s, wee obtain that $g$ has at least a root in $]-d_k,-d_{k-1}[,\cdots$, in $]-d_{p+1},-d_p[$, in $]-d_{p-1},-d_{p-2}[,\cdots$, in $]-d_2,-d_1[$; moreover, we know that $g$ has at least a root $>-d_1$ and thus, we have our $k-1$ roots and we are done.

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  • $\begingroup$ Thanks. If I understood correctly, In case $2$ you meant to say the following?: $\lim_{x\rightarrow +\infty}f(x)=1+$, so $f(x) > 1$ for $x$ large enough. On the other hand $f(\sup_i(-d_i))=0$, so by the intermediate value theorem, there exist an $m$ s.t $f(m)=1$. Since $f$ is monotone this $m$ is unique. So, to summarize: If there exist a $d_i<0$, then there exist a unique solution where $m>0$. Otherwise (when all the $d_i \ge 0$, the only solution is when $m=0$. So, in any case (no matter what $D$ is) there is a unique solution, right? $\endgroup$ – Asaf Shachar May 3 '17 at 13:45
  • $\begingroup$ BTW, I agree with you this is not related so much to linear algebra (so I removed the tag). Your idea to "separate" the question to finding the determinant and the "elements" separately is very nice. Thanks again for your effort. $\endgroup$ – Asaf Shachar May 3 '17 at 16:23
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I am merely clarifying some points regarding loup blanc's answer:

Denote $D=\diag(d_i),\Sigma=\diag(\sigma_i),m=(\Pi_i{\sigma_i}^2)^{1/d}$, and suppose $\sigma$ is a solution to the equation fir $D$. Then, necessarily $$m\geq \sup_i(-d_i) \, \, \text{and} \, \, \sigma_i=\sqrt{m+d_i} \tag{1}.$$

Note that in any case $m \ge 0$.

A first observation:

$m=0 \iff d_i \ge 0$ for all $i$.

Proof:

$m=0 \Rightarrow d_i \ge 0$: Obvious from equation $(1)$.

$d_i\ge 0 \Rightarrow m= 0$:

Suppose by contradiction that $m > 0$. The assumption $d_i\ge 0$ implies $\sigma_i^2 \ge m$, so $m^d=\Pi_{i=1}^d \sigma_i^2 \ge m^d>0$, but this can happen only if $\sigma_i=m$ for all $i$, and then $\Sigma$ is conformal, a contradiction.

Back to the main proof:

We want to prove that for all $D$ satisfying the conditions stated in the question, there is a unique solution. We separate into cases"

(1) $d_i \ge 0$ for all $i$.

The n $m=0$, so equation $(1)$ reduces to $\sigma_i=\sqrt{d_i}$.

(2) There exist an $i$ such that $d_i < 0$.

Then loup blanc's argument shows that there exist a solution. Let's show there is a unique solution:

We need to show the function $$f:x\in[\sup_i(-d_i),+\infty]\rightarrow \Pi_j(1+d_j/x) $$ obtain the value $1$ at most at one point.

We already know that $f(\sup(-d_i))=0, \lim_{x \to \infty}f(x)=1$ and that $f(x) > 1$ for $x$ sufficiently large. Also, it is easy to see $f'(\sup(-d_i))>0$. Hence, if we will show $f'(x)$ has only one zero than this would imply $f$ obtaines the vlaue $1$ exactly at one point.

For $x > \sup(-d_i)$, $$ f'(x)=-\sum_j \frac{d_j}{x^2}\Pi_{i\neq j}(1+\frac{d_i}{x})=-\sum_j \frac{d_j}{x^2}\Pi_{i}(1+\frac{d_i}{x})\frac{1}{1+\frac{d_j}{x}}=-\sum_j \frac{d_j}{x^2}f(x)\frac{1}{1+\frac{d_j}{x}},$$

so $$f'(x)=-\frac{f(x)}{x^2}\sum_j d_j\frac{1}{\frac{x+d_j}{x}}=-\frac{f(x)}{x^2}\sum_j \frac{d_jx}{x+d_j}.$$ Hence, for $x > \sup(-d_i)$, $f'(x)=0$ if and only if $h(x):=\sum_j \frac{d_jx}{x+d_j}=0$. Assume by contradiction that $h$ has two zeroes. Than, $h'(x)=0$ for some $x$ but

$$h'(x)=\sum_j \frac{d_j^2}{(x+d_j)^2}$$ which is impossible.


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