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Suppose I have a matrix $X\in \mathbb{R}^{n\times n}$, such that

  1. $X$ is symmetric
  2. Do not know the rank

I want design a matrix function $f(X,Q)\in \mathbb{R}^{n\times n}$ with $Q = qq^T$ and $q\in \mathbb{R}^n$, such that

  1. $f(X,Q)$ is symmetric
  2. $\text{tr} (f(X,Q))=0$
  3. rank$(f(X,Q))$ is $2$.

For

  1. we can design $QX+XQ$
  2. I only know the property $\Omega Q + Q\Omega^T$ with $\Omega$ a skew-symmetric matrix (since tr$(Q\Omega)=q^T\Omega q=0$.
  3. Choose $f(X,Q)=QX+XQ$. The basis for its column space is $\{q, Xq\}$, suppose $Xq \neq q$.

However, I have no idea how to combine all of them. Can anyone help me this?

Sincerely appreciate this help.

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    $\begingroup$ (3) is not necessarily true for your example. The rank could be $0$ or $1$. $\endgroup$
    – Robert Israel
    Commented Jan 26, 2021 at 19:01
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    $\begingroup$ The constant function f(X, Q) = diag(1, -1, 0, ..., 0) satisfies all properties, but it is most likely not what you want, right? $\endgroup$
    – A.Z.
    Commented Jan 26, 2021 at 22:33
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    $\begingroup$ @A.Z. The matrix function here I mean, for example $f(X,Q) = 2X+3Q+4QX$. $\endgroup$
    – Denny
    Commented Jan 27, 2021 at 11:19
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    $\begingroup$ $Xq \ne q$ is not enough. You need $q$ and $Xq$ to be linearly independent. $\endgroup$
    – Robert Israel
    Commented Jan 27, 2021 at 16:54
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    $\begingroup$ Do you want your function to be a matrix polynomial in $X$ and $Q$, with constant coefficients? $\endgroup$
    – Will Sawin
    Commented Jan 31, 2021 at 15:30

1 Answer 1

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There is no such function

Assumption on the Function space

I assume that $f$ has to be a polynomial in the variables $X$ and $Q$, i.e., $f$ is a sum of terms of the form

$$ a_k X^{k_1} Q^{k_2} X^{k_3} \dots Q^{k_j} $$

where $a_k$ is a real/complex number.

$$ f(X,Q) = \sum_{i=0}^\infty \sum_{\substack{k \text{ is an integer} \\\\ \text{partition of } i}} a_{k} X^{k_1} Q^{k_2} X^{k_3} \dots Q^{k_j} $$

If this is not the class of functions you had in mind, could you specify further?

The proof

As there are no restrictions on $X$ and $Q$ the zero matrices are possible choices for $X$ and $Q$ (choose $q$ to be the zero vector for $Q$). Then all terms of $f$ which involve at least one $X$ or $Q$ are 0. $$ f(0,0) = a_0 X^0 Q^0 = a_0 \mathbb{I} $$ with the identity matrix $\mathbb{I}$. Thus except for the case of $n=2$, $a_0 \neq 0$ the matrix $f(0,0)$ never has rank 2, however in this case $f(0,0)$ does not have trace $0$.

Alternatives

There are several ways to modify the assumptions to get this to work:

  • Restrict the set of matrices $X$ and $Q$ for which the requirements have to hold. Especially, exclude the zero matrices.
  • Extend the set of possible functions, to, for example, include the constant function $f(X,Q) = \operatorname{diag}(1, -1, 0, \dots, 0)$.
  • Drop one of the requirements.
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  • $\begingroup$ I just come up with such possible function. Not sure if it is correct. $f(X,Q) = XQ-QX$. It is symmetric, rank 2 with $\{Xq, q\}$ if $Xq, \, q$ are L. I. And trace of $f(X,Q)$ is zero since tr$(XQ) = q^TXq$, so tr$(f)=q^TXq-q^TXq=0$. $\endgroup$
    – Denny
    Commented Jan 29, 2021 at 16:53
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    $\begingroup$ If $q$ is the zero vector then $Xq$ and $q$ are not linear independent. Is $q=0$ allowed for your problem? $\endgroup$
    – A.Z.
    Commented Jan 31, 2021 at 17:26

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