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Consider two skew-adjoint matrices $A$ and $A'$, i.e. $A^*=-A$ and $A'^*=-A'$. It is well-known that

$e^{-tA}$ and $e^{-tA'}$ are unitary operators.

I would like to know:

Is it true that $\sup_{t \in \mathbb{R}} \Vert e^{-tA}-e^{-tA'} \Vert = 2(1-\delta_{A,A'})?$

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  • $\begingroup$ What is $\delta_{A, A'}$? $\endgroup$
    – Nik Weaver
    Commented May 25, 2020 at 19:45
  • $\begingroup$ the kronecker delta that is 1 if $A=A'$ and zero otherwise. $\endgroup$
    – Kung Yao
    Commented May 25, 2020 at 19:47
  • $\begingroup$ I see, thank you. $\endgroup$
    – Nik Weaver
    Commented May 25, 2020 at 19:50
  • $\begingroup$ So the question is: if two strongly continuous unitary groups are not identical, then must their pointwise distance be 2? $\endgroup$
    – Nik Weaver
    Commented May 25, 2020 at 19:51
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    $\begingroup$ Please don't delete your question after it's been given an answer, this is disrespectful. $\endgroup$
    – YCor
    Commented May 25, 2020 at 20:57

1 Answer 1

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This is not true in general. E.g., let $$A:=\left( \begin{array}{cc} -i & 0 \\ 0 & 0 \\ \end{array} \right),\quad A':=\frac i2\, \left( \begin{array}{cc} -1 & 1 \\ 1 & -1 \\ \end{array} \right).$$ Then, for real $t$, $$e^{-tA}=\left( \begin{array}{cc} e^{i t} & 0 \\ 0 & 1 \\ \end{array} \right),\quad e^{-tA'}=\frac12\,\left( \begin{array}{cc} 1+e^{i t} & 1-e^{i t} \\ 1-e^{i t} & 1+e^{i t} \\ \end{array} \right),$$ and $$\|e^{-tA}-e^{-tA'}\|=\sqrt{1-\cos t}, $$ so that $$\sup_{t\in\mathbb R}\|e^{-tA}-e^{-tA'}\|=\sqrt2\ne2(1-\delta_{A,A'}).$$

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