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I see this question and I have a new question which is a special form of it: what can we say about a linear function which can commute with every projections?

fateme jl
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All such maps are some multiple of the identity.

Let $p$ be a projection onto a subspace $V$. If $v\in V$ then $p(v)=v$. So if $f$ commutes with $p$ and $v\in V$ we have $f(v)=f(p(v))=p(f(v))\in V$, so $f$ sends $V$ to a subspace of $V$. So if $f$ commutes with every projector it sends every subspace to a subspace of that subspace.

So by looking at the one dimensional subspaces we see that for every vector $v$ we have $f(v)=\lambda v$ for some $\lambda$. I claim that in fact this must be the same $\lambda$ for each vector. Suppose $f(v)=\lambda v$ and $f(w)=\mu w$. Then

$$f(v+w)=\lambda v+\mu w,$$ and this must equal $\nu(v+w)$ for some $\nu$. So $(\lambda-\nu)v=(\nu-\mu)w$. If $v$ and $w$ point in different directions this can only happen if both sides are $0$, so $\lambda=\nu=\mu$.

Hence $f$ is $\lambda$ times the identity for some $\lambda$.

Oscar Cunningham
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  • I don't know why here we have : f(v)=f(p(v)) – fateme jl Apr 14 '17 at 10:19
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    @fatemejl $v=p(v)$ because $v\in V$ and $p$ is a projector that projects onto $V$. So $f(v)=f(p(v))$. – Oscar Cunningham Apr 14 '17 at 10:24
  • Can you explain more about the equality? I'm a bit confused. I see no necessity for holding this equality for a projector like p. thanks for your help – fateme jl Apr 14 '17 at 10:37
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    The definition of projector requires that $p^2=p$. If $V$ is the image of $p$ and $v\in V$ then $v=p(v')$ for some $v'$. So $p(v)=p(p(v'))=p(v')=v$. – Oscar Cunningham Apr 14 '17 at 10:42
  • in last part of the proof that you claim this multiplicity is unique, what will happen if we choose v and w from different subspaces? – fateme jl Apr 15 '17 at 05:14
  • If they point in different directions then my proof works because then the only way to have $(\lambda -\nu)v=(\nu-\mu)w$ is to have $(\lambda -\nu)v=0$ and $(\nu-\mu)w=0$. If $v$ and $w$ point in the same direction then we can write $w=\alpha v$ for some number $\alpha$ and then $f(w)=f(\alpha v)=\alpha f(v)=\alpha\lambda v=\lambda\alpha v=\lambda w$, so that case works too. – Oscar Cunningham Apr 15 '17 at 07:25
  • I mean if you choose two different vector with different directions, how you claim that f(v+w) is the multiple of v+w? – fateme jl Apr 16 '17 at 20:46
  • we know that this holds in the subspaces, not all the space. – fateme jl Apr 16 '17 at 20:47
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    We know that $f$ takes every subspace into itself. So in particular the one-dimensional subspace spanned by $v+w$ gets taken into itself. So $v+w$ gets mapped to some multiple of itself. – Oscar Cunningham Apr 16 '17 at 21:41
  • can you explain why p(f(v)) is a member of V? how we know that f(v) is also a member of V? – fateme jl May 20 '17 at 17:58
  • The image of $p$ is $V$, so $p$ of anything is in $V$. In particular $p(f(v))$ is in $V$. – Oscar Cunningham May 21 '17 at 08:47