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A simple question that I was pondering on while examining some algorithms that work similarly for positive definite and nonnegative matrices.

Let $\mathcal{H}$ be the space of (let's say for now $2\times 2$) Hermitian positive-definite matrices. Let $\mathcal{P}$ be the space of entrywise nonnegative matrices of the same dimension.

One can define a linear map $\Phi:\mathcal{H} \to \mathcal{P}$ in many ways; for instance, $$ \begin{bmatrix} a & b+id\\b-id & c \end{bmatrix} \mapsto \begin{bmatrix} a & a+c+b\\a+c+d & c \end{bmatrix}. $$ Is there a such a linear map that preserves matrix squaring, i.e., a map $\Phi: \mathcal{H} \to \mathcal{P}$ such that

  1. $\Phi(tH)=t\Phi(H)$ for each $t >0 $,
  2. $\Phi(H+K)=\Phi(H)+\Phi(K)$,
  3. $\Phi(H^2)=\Phi(H)^2$ ?
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  • $\begingroup$ Requirement 3 almost requires the map $\Phi$ to be multiplicative, which means that only maps of the form $\Phi(X) = X \otimes I$ etc. might fulfill the condition, so the answer to your question might be "no"... $\endgroup$
    – Suvrit
    Jul 14, 2013 at 17:44
  • $\begingroup$ Hi Suvrit! What do you mean by multiplicative in this context? The product of two Hermitian matrix is not Hermitian in general. Also, if one drops the positivity requirement, if I have made no mistakes then $\begin{bmatrix}a & b+id\\b-id & c\end{bmatrix} \mapsto \begin{bmatrix}a & b+d \\ b-d & c\end{bmatrix}$ satisfies 1.--3. and is not the trivial map. $\endgroup$ Jul 14, 2013 at 19:03
  • $\begingroup$ Do you mean "nonnegative semidefinite" when you say "positive definite"? I'm asking because I see $t\geq 0$ in 1., rather than $t > 0$. $\endgroup$ Jul 14, 2013 at 21:56
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    $\begingroup$ An obvious (and useless) example is $\Phi(H) = 0$. What other properties should $\Phi$ have (injectivity, for example)? $\endgroup$ Jul 15, 2013 at 15:42
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    $\begingroup$ Maybe I'm missing something, but I don't see how your example $\begin{bmatrix}a & b+id\\b-id & c\end{bmatrix} \mapsto \begin{bmatrix}a & b+d \\ b-d & c\end{bmatrix}$ satisfies $3$. When I square the left side, I get the top left element $a^2+b^2\color{red}{+}d^2$, but when I square the right side, I get $a^2+b^2\color{red}{-}d^2$ and, by your map, they should be the same. What am I missing? $\endgroup$ Jul 15, 2013 at 15:46

1 Answer 1

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This is not an answer, but a collection of observations and ideas, in hopes that it helps someone to get more.

Let us ask a bit more of our function:

  1. If $AB$ is positive semidefinite, then $\Phi(AB) = \Phi(A)\Phi(B)$.
  2. $\Phi(H)$ is nonsingular for at least one $H \in \mathcal{H}$. (I think this is quite a reasonable request to make)

Notice that the first of these two requests also means that if $A,B$ commute, then $\Phi(A),\Phi(B)$ commute as well. Using this, we easily see that

$$t\Phi(H) = \Phi(t{\rm I}H) = \Phi(t{\rm I})\Phi(H), \quad \text{for all $H$}.$$

We pick $H$ such that $\Phi(H)$ is nonsingular and get

$$\Phi(t{\rm I}) = t{\rm I}, \quad \text{for all $t \ge 0$}.$$

This also draws a trivial conclusion that

$$\Phi(H^{-1}) = \Phi(H)^{-1}.$$

If $A$ is a matrix whose elements are all equal to $1$, then

$$nf(A) = f(nA) = f(A^2) = f(A)^2.$$

Unfortunately, all-ones matrices are not the only ones with such property, so I cannot get anything more from this (I was hoping to get some insight on $\Phi(A)$, at least for such special $A$).

The preservation of commutativity gave me the following idea. If $A,B \in \mathcal{H}$ commute, then there exist a unitary matrix $U$ and nonnegative diagonal matrices $\Lambda_A,\Lambda_B$ such that $$A = U \Lambda_A U^*, \quad B = U \Lambda_B U^*.$$ But, since this means that $\Phi(A),\Phi(B)$ also commute, there exist a unitary matrix $V$ and upper triangular matrices $T_A,T_B$ such that $$\Phi(A) = V T_A V^*, \quad \Phi(B) = V T_B V^*.$$ So, it may make sense to define $\Phi(X)$ as a map $(U,\Lambda) \mapsto (V, T)$, where $X = U \Lambda U^*$ for $U$ unitary and $\Lambda$ nonnegative diagonal, and $\Phi(X) = V T V^*$ for $V$ unitary and $T$ uppertriangular.

The problems with this approach are:

  1. defining such map to guarantee that $\Phi(X)$ is nonnegative,
  2. property 2 gets hard to prove for noncommutative matrices.

The first problem might be solved by changing the requirements on $V$. Maybe it can be some nonsingular matrix with nonnegative elements (so, not unitary). The question of how to get such matrix remains.

I'm currently at loss for additional ideas. I hope these will help at least a bit.

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  • $\begingroup$ I think that the multiplicativity restriction that you mention pretty much forces $\Phi$ to be of the form $X \otimes I$ as I mentioned in my comment to Federico....so.... $\endgroup$
    – Suvrit
    Jul 16, 2013 at 13:38
  • $\begingroup$ "No" was my hunch as well. However, I've managed to find no solid proof for it, so I went looking for such a map, in hopes to either get something useful or get a contradiction (all-ones matrix was one of such tries). $\endgroup$ Jul 16, 2013 at 14:14

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