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Inspired by this question.

It is known that for the matrix $T_n \in \mathcal{M}_n$ (the space of real-valued $n \times n$ matrices) defined by \begin{equation*} (T_n)_{ij} = \begin{cases} 1 & i \geq j \\ 0 & i < j \end{cases} \end{equation*}

and arbitrary $A \in \mathcal{M}_n$, that $\|T_n \circ A\| \leq \frac{\ln n}{\pi}\|A\|$, where $\circ$ represents Hadamard/Schur multiplication and $\|\cdot\|$ represents the spectral norm.

My question is this:

Let $B \in \mathcal{M}_n$ be symmetric. Is the above inequality still best possible, or can it be improved? To be more exact, is there a sequence of symmetric $n \times n$ matrices $B_n$ such that $\|T_n \circ B_n\| \approx \|B_n\|\ln n$ ?

Try as I might, I can find no results in the literature about this (though I can find many about the norm of Hadamard multiplication with $T_n$ acting on $\mathcal{M}_n$) and I cannot seem to come up with an example of a suitable $\{B_n\}$ myself.

Any help would be much appreciated, even if it is to recommend a resource or reference. Thanks so much!

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  • $\begingroup$ Since you say symmetric, I am guessing that your matrices are real-valued rather than complex-valued? $\endgroup$
    – Yemon Choi
    Apr 8, 2021 at 22:45
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    $\begingroup$ This is answered in the question you linked to: the best constant is still $\sim\log n$. (What you call $T\circ B$ corresponds to $D$ there, and of course $D+D^t$ is symmetric and can be your $B$.) $\endgroup$
    – Christian Remling
    Apr 9, 2021 at 1:50
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    $\begingroup$ This is untrue; the triangular cut used in Kwapien and Pelczynski is for everything below the main cross-diagonal, not the main diagonal of the matrix. Their result doesn't apply here, unfortunately, as you would have to "rotate" the matrix somewhat and in doing that you would get an anti-symmetric matrix. $\endgroup$
    – SAWblade
    Apr 9, 2021 at 13:27
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    $\begingroup$ I remember trying symmetric random matrices with Gaussian N(0,1) entries; for these the upper triangular based truncation had $\frac{1}{\pi}\log n$ type of bound....cannot locate my notes about that right now; at least empirically this can be checked and perhaps proved subsequently directly. $\endgroup$
    – Suvrit
    Apr 9, 2021 at 16:12
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    $\begingroup$ @SAWblade You are right, I read Kwapien-Pelczynski too fast. But using that $iA$ is hermitian when $A$ is skew-symmetric, you can pass from skew-symmetric to symmetric and conversely. See my answer. $\endgroup$
    – Mikael de la Salle
    Apr 11, 2021 at 20:39

1 Answer 1

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You seem to be accepting that the norm of the triangular projection restrited to skew-symmetric matrices is of order $\log n$. There is an easy trick to deduce that the same is true on symmetric matrices. Operator algebraists like it a lot and call it the $2$-by-$2$ matrix trick.

The trick is: if $A \in M_n$ is skew-symmetric, then $A'=A\otimes J =A \otimes \begin{pmatrix} 0 & 1 \\-1&0\end{pmatrix}$ is symmetric. If you choose the correct identification between $M_n \otimes M_2$ and $M_{2n}$, then the triangular projection of $A'$ is almost $(T_n \circ A)\otimes J$ (it is exactly $(T_n \circ A)\otimes J - (I_n \circ A) \otimes \begin{pmatrix}0&0\\-1&0\end{pmatrix}$). This implies that $$\|T_{2n}\circ A'\| \geq \|T_n \circ A\| - \|A\|,$$ and that the norm of $T_{2n}$ restricted to symmetric matrices is larger that (the norm of $T_n$ restricted to skew-symmetric matrices) minus $1$, QED.

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  • $\begingroup$ Could you please elaborate on "the correct identification between $M_n \otimes M_2$ and $M_{2n}$"? $\endgroup$
    – SAWblade
    Apr 12, 2021 at 16:53
  • $\begingroup$ I ask because I cannot see how $T_{2n} \circ A' = (T_n \circ A) \otimes J - (I_n \circ A) \otimes \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix}$. $\endgroup$
    – SAWblade
    Apr 12, 2021 at 17:38
  • $\begingroup$ @SAWblade Order the elements of $\{1,\dots,n\} \times\{1,2\}$ as follows: $(1,1),(1,2),(2,1),(2,2),(3,1),\dots,(n,1),(n,2)$. $\endgroup$
    – Mikael de la Salle
    Apr 13, 2021 at 7:38
  • $\begingroup$ ... in that way, if $0\leq i,j\leq n-1$ and $1\leq \alpha,\beta\leq 2$, the entry $(2i+\alpha,2j+\beta)$ of $A'$ is $A_{i+1,j+1}J_{\alpha,\beta}$. $\endgroup$
    – Mikael de la Salle
    Apr 13, 2021 at 7:41
  • $\begingroup$ I believe this is incorrect as $I_n \circ A$ would be the zero matrix, and also $(T_n \circ A)\otimes J - A \otimes \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix}$ is equal to the symmetric matrix $B$ where the upper right corner is $T_n \circ A$, not $T_{2n} \circ A'$. $\endgroup$
    – SAWblade
    Apr 14, 2021 at 22:10

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