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Toni Mhax
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I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$. The only thing is that it may(not) hold for $n=3$.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$. The only thing is that it may(not) hold for $n=3$.

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Toni Mhax
  • 785
  • 5
  • 13

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$ or $n$ odd.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$ or $n$ odd.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$.

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Toni Mhax
  • 785
  • 5
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I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$ or $n$ odd.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

I came across the following while doing some related proof; It seems easy to prove. $\quad$
We are in ${\mathbb{M}}_n(\mathbb{C})$, $n>1$:

$1$) Given a unitary $n\times n$ matrix $U$, there is some permutation of the columns such that the modulus of each entry on the diagonal of the resulting matrix is $\le \dfrac{\sqrt{2}}{2}$, $\quad$

And more difficult:

$2$) There are no $n\times n$ unitary matrices $U$ of diagonal $D$ with all entries satisfying $|d_{i,i}|>\dfrac{1}{\sqrt{n-1}}$ for all $i$, unless $U$ is a direct sum of unitaries $U_k$, (up to a permutation congruence) $U_k\in {\mathbb{M}}_k$, $k<n$. I am searching for a proof or related facts.
Thanks.

Edit: for $2$ and $n=4$, we can take $\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}$ The only thing is that it may hold for $n=3$ or $n$ odd.

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Toni Mhax
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