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Two projection operators may not be commutative, with an explicit example.
A projection operator is a linear mapping from a linear space X to X that then satisfies $p^2=p$ Then in a geometrical point of view, looking at the linear space, any element in $X$ mapping under $p$ only has two possibilities, one is $px=x$, and under the action of $p$ image of $x$ is himself. The second possibility, x under the action of $p$ is not itself but another element $y$, then $p$ in action, on $y$ should also be equal to $y$ because we know that $p^2x=px = y$ so $py=p^2x=px = y$. Well…
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分析中很多问题都可以用取平均来解决,但是能用取平均解决的问题,本质上是trivial的。 Lonely Runner conjecture本质上是一个不能用取平均解决的问题,整个框架就不对, 可以看成一个粒子扩散的问题,到地方那看这个就可以看出一个例子扩散的问题,但是它是非常非常简单的情形,在这个情形下,整个全空间是$S_1$或者$Z_p$。