0

I'm trying to prove that if $V$ is a vector space and a linear operator $A: V \rightarrow V$ commutes with all $B \in \mathcal{L}(V,V)$ then $A= \lambda I$ and I need help. Here is what I have so far.

Since $A$ and $B$ commute then all the eigenspaces of $A$ are invariant under $B$. So $\forall v \in E_\lambda, A \in E_\lambda$. Thus, $ABv= \lambda Bv$ and since the two operators commute, it follows that $BAv= \lambda Bv= B(\lambda v)$. Hence, $Av=\lambda v$. Therefore we want to show that this is true for all $v \in V$ i.e, $E_\lambda=V$.

I tried playing around with some different options of $B$ such as a projection operator onto an arbitrary subspace or an automorphism, but I'm having trouble formalizing it into a proof and need help.

sloemz
  • 1
  • 3
    Do you mean "if $A:V\to V$ commutes with all $B\in\mathcal{L}(V,V),$ then $\color{red}{A}=\lambda I$? – Gonçalo Dec 22 '24 at 03:51
  • 1
    Your idea of using commuting operators is interesting. But why didn't you use it this way?: $A, B$ commute so all eigenspaces of $B$ are invariant under $A$. Let's choose operators $B$ with a 1-dimensional eigenspace (generated by some arbitrary vector $v$) then $v$ is also an eigenvector of $A$. Hence every nonzero vector is an eigenvector of $A$. Can you go on from here? – Duong Ngo Dec 22 '24 at 04:02
  • @DuongNgo a simpler idea of choosing $\mathscr{B}$ may be since $\mathscr{A},\mathscr{B}$ commute, then $\operatorname{im}\mathscr{B}$ is invariant under $\mathscr{A}$. – syqwq Dec 22 '24 at 06:24

2 Answers2

1

Hint

try prove:

$V$ is a vector space and $\mathscr{A}:V \to V$ is a linear operator. If $\lambda_1 \neq \lambda_2$ are 2 eigenvalues of $\mathscr{A}$, and $\xi_i$ is an eigenvector of $\mathscr{A}$ corresponding to $\lambda_i$, $i=1,2$. Then, $\xi_1+\xi_2$ is not an eigenvector of $\mathscr{A}$.

syqwq
  • 111
0

It's been a while since I thought about this, but I read the comments and Duong Ngo gave me an idea of another way to think about it. (Sorry I don't know how to reference someone). Let me know if I'm correct.

Let $A:V \rightarrow V$ be a linear operator and let $ B \in \mathcal{L}(V,V)$. Since A and B commute, all the eigenspaces of A are invariant under B. Choose B to be an arbitrary projection onto an arbitrary subspace $X$. Because the eigen spaces of $A$ are invariant under $B$, $Bv$ is an eigenvector of A for all $v\in V$. Since $X$ is an arbitrary subspace, then any vector in $V$ must be an eigenvector of $A$. Thus, $A=\lambda I$ .

sloemz
  • 1