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Let $A,B \in Mat_{n,n}(\mathbb C)$, $AB=BA$ and $\lambda$ is an eigenvalue for $B$.

I've shown that $A$ is a linear operator on $E_B(\lambda)$, that is if $v \in E_B(\lambda)$ then $Av \in E_B(\lambda)$.

Now I must show that $E_B(\lambda)$ contains an eigenvector for $A$.

I've tried something like:

Let $u = Av$.

$$\lambda u = Bu = BAv = ABv = A(\lambda v) $$

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    $\begingroup$ What does $E_B(\lambda)$ mean? $\endgroup$ – Gerry Myerson Apr 30 '14 at 9:49
  • $\begingroup$ By the way, from your equation, you get $B(Av)=\lambda(Av)$. Is that by any chance what you want? $\endgroup$ – Gerry Myerson Apr 30 '14 at 9:50
  • $\begingroup$ $E_B(\lambda)$ denote the eigenspace corresponding to the eigenvalue $\lambda$. What I want is a vector $v$ in this eigenspace such that $\lambda v = A v$ . $\endgroup$ – Shuzheng Apr 30 '14 at 9:52
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If $x\in E_B(\lambda)$ then $B(A(x))=A(B(x))=A(\lambda x)=\lambda A(x)$, that is $A(x)\in E_B(\lambda)$. This proves that $A$ defines an operator $\tilde A$ from $E_B(\lambda)$ to $E_B(\lambda)$ by $\tilde A(x)=A(x)$. Now, any eigenvector $v$ of $\tilde A$ is an eigenvector of $A$ that belongs to $E_B(\lambda)$.

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  • $\begingroup$ Thanks, but have you proven that there exist an eigenvector for the operator defined by $A$ in $E_B(\lambda)$ ? $\endgroup$ – Shuzheng Apr 30 '14 at 13:03
  • $\begingroup$ Since we're working over the complex numbers, every operator has an eigenvector; in particular, $\tilde A$ has an eigenvector, and that eigenvector is both an eigenvector of $A$ and is in the eigenspace of $B$. $\endgroup$ – Gerry Myerson Apr 30 '14 at 13:17
  • $\begingroup$ I understand $A$ must have an eigenvector because of $p_A(\lambda)$ and we are in $\mathbb C$. But what about the linear operator induced by $A$ ? Here we are dealing in a subspace of $\mathbb C$ (the eigenspace), and how do I show that it has an eigenvector in $E_B(\lambda)$ ? - By finding the corresponding matrix or ? $\endgroup$ – Shuzheng Apr 30 '14 at 13:21
  • $\begingroup$ The matrix of $\tilde A$ as operator on $E_B(\lambda)$ has an eigenvector, this is it. $\endgroup$ – Omran Kouba Apr 30 '14 at 13:35

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