Suppose $S$ and $T$ are two linear maps from $V$ to $V$. let $v$ be an non-zero eigen vector corresponding to some eigen value $\lambda$ of $TS$. If $S(v)=0$ then $\lambda =0$. So, TS has only zero eigen value. Is it also true that ST also has zero eigen value under the condition that $S(v)=0$?

  • $\begingroup$ if $T(v)=v$ it certainly is the case since $ST(v)=S(T(v))=S(v)=0$ $\endgroup$ – Eleven-Eleven Oct 9 '13 at 9:39
  • $\begingroup$ If $TS$ has a zero eigenvalue does not generally mean that $ST$ has one as well. All you can say is that $TS$ and $ST$ have the same nonzero part of their spectrum. $\endgroup$ – Algebraic Pavel Oct 9 '13 at 9:41
  • $\begingroup$ If $S(v)=0$ then $S$ has non-trivial kernel. So, $S$ is not one-one. So $ST$ is not one-one and so it has non-trivial kernel which implies $ST$ has zero eigen value. $\endgroup$ – user96000 Oct 9 '13 at 10:01
  • $\begingroup$ You mean if $S(v)=0$ for some $v$ then $ST$ has a non-trivial kernel? Not necessarily if $\mathrm{Im}(T)\cap\mathrm{Ker}(S)=\{0\}$. $\endgroup$ – Algebraic Pavel Oct 9 '13 at 10:04

In this case, the answer is yes, because $\operatorname{rank} ST = \operatorname{rank} TS$ and your operators go from $V$ to $V$ (i.e., their matrix representations are square matrices). However, an eigenvector corresponding to the eigenvalue $\lambda = 0$ need not be the same for $ST$ and $TS$.

By the way, eigenvalue and eigenvector are written as single words.


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