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  1. Let $G$ be a group of "diagonal sign matrices $(g)$", i.e. a diagonal matrix with diagonal entries in $\{-1,1\}$. So $G$ is of cardinality $2^n$. For example: $$g = \begin{bmatrix}1 & 0 &0 \\0 & -1 & 0 \\0 & 0 & 1 \end{bmatrix}$$

  2. Define $$P(X) = \frac{1}{2^n}\sum_{g\in G} gXg^T$$ with $X\in \mathbf{R}^{n\times n}$

My question: How to show $P$ is self-adjont?

So, obviously we have to show $P^T=P$:

$$P^T = \big(\frac{1}{2^n}\sum_{g\in G} gXg^T\big)^T = \frac{1}{2^n}\sum_{g\in G} \big(gXg^T\big)^T=\frac{1}{2^n}\sum_{g\in G} gX^Tg^T$$

How to go a step further?

http://arxiv.org/abs/1403.4914 (p.1326)

enter image description here

sleeve chen
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  • What exactly is $X$ here? I.e., what space is $P$ operating on? – Mark Sep 08 '16 at 00:45
  • @Mark I correct my article. $P$ just operate on $X$ by multiplication of $g$ and summation over all elements in $G$ and average (divided by $2^n$). $X\in R^{n\times n}$, so $X$ is arbitrary square real matrix with dimension $n$ – sleeve chen Sep 08 '16 at 00:52

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Choose some $h\in G$ and fix $X$. Then \begin{equation} hP(X)h^T = \frac{1}{2^n} \sum_{g\in G} hgXg^Th^T = \frac{1}{2^n} \sum_{g\in G} (hg)X(hg)^T = P(X). \end{equation} So we have \begin{equation} P(P(X)) = \frac{1}{2^n}\sum_{g\in G} gP(X)g^T = \frac{1}{2^n}\sum_{g\in G} P(X) = P(X), \end{equation} meaning that $P$ is idempotent. Some similar trickery will show that $PP^T = P^T$ and that $P^T P=P^T$. The fact that $PP^T = P^TP$ means that $P$ is a normal operator, and you can find arguments (e.g., here or here) showing that all normal, idempotent operators are self-adjoint.

Community
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211792
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  • How could you say $P$ is idempotent? An idempotent matrix $M$ is defined as: $MM = M$. $P(P(X)) = P(X)$ looks like composition rule? – sleeve chen Sep 08 '16 at 01:38
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    Idempotence isn't just for matrices: a function $f$ is called idempotent if $f=f\circ f$ (when this composition makes sense). – 211792 Sep 08 '16 at 01:41
  • @AustinC could you give more details about how you derive $PP^T=P^T$? – stewbasic Sep 08 '16 at 01:44
  • @stewbasic As with $P$, we can show that $hP^T(X)h^T=P^T(X)$ for every $h\in G$ and every $X$. Then we can compute $P(P^T(X))=P^T(X)$ in the same way that $P(P(X))$ is computed in the second line above. – 211792 Sep 08 '16 at 03:26
  • @AustinC This requires first describing $P^T(X)$, right? As far as I can see, the easiest way to do so is to show that $\langle P(X),Y\rangle=\langle X,P(Y)\rangle$ so $P^T=P$, but then the question is already answered. – stewbasic Sep 08 '16 at 04:17
  • @stewbasic $P^T(X)$ is given in the original post (though it's written as just $P^T$). – 211792 Sep 08 '16 at 13:33
  • @AustinC The expression in the original post is for $P(X)^T$, which is not in general the same as $P^T(X)$ (it happens to be in this case, but that is not clear a priori). – stewbasic Sep 08 '16 at 21:41
  • Good catch. With this in mind, I don't immediately see how to patch up my solution, so I'll delete it shortly. Thanks for catching my error! – 211792 Sep 08 '16 at 23:51
  • @AustinC I provide the original article of my problem. $P(X)$ is a function, if fact $P(X)$ is a diagonal matrix whatever $X$ is. In fact you do not need to delete your answer, your answer provides a good way to thinking, – sleeve chen Sep 09 '16 at 00:50
  • The answer can't be deleted since it's been accepted, but I still think it's not quite right, as per stewbasic's comment. It should be noted that the proof in the linked paper doesn't match the snippet given above. – 211792 Sep 09 '16 at 02:20