Given transformations $S, T$ on $V$ (dim($V$)=$n$), and $S^2=0=T^2, ST=-TS+I$, where I is the identity transformation. Find all eigenvalues of $S, T$ and their corresponding algebraic multiplicities and geometric multiplicities.
From the conditions $S^2=T^2=0$, $\forall x \in V$, it is easy to find that $$S(Sx)=0\\ Sx\in Ker(S)$$
Then $\forall y\in Ker(S)$, we have $Sy=0$, then $y=S(2Ty)$. Finally we have $Ker(S)=Im(S)=Sker(T)$, so is $T$.
Hence for eigenvalues of $S$, I can only find a $0$ by directly looking, and I guess they are the specific number like $0, 1$. For algebraic and geometric multiplicities, I cannot get them in a general way, it seems hard to get the characteristic polynomial.
Can someone help me? Thank you.
Best Answer
By the given conditions, we see that $ST$ and $TS$ are idempotent. Since $ST$ and $TS$ also share the same spectrum, they are projections of the same rank $r$.
Let $X=STV$ and $Y=TSV$. Then $\dim X=\dim Y=r$. As $STY=0$, we see that $X\cap Y=0$. Since $ST+TS=I$, we infer that $V=X\oplus Y$ and $n=2r$.
The given conditions also imply that $S=STS$. Therefore $SV=STSV\subseteq STV\subseteq SV$. Hence $SV=STV$ and $\operatorname{rank}(S)=\operatorname{rank}(ST)=r$.
Since $S$ is nilpotent, all its eigenvalues are zero. The geometric multiplicity of the zero eigenvalue is the thus nullity of $S$, which, by the rank-nullity theorem, is equal to $n-r=r$. The same conclusion can be drawn for $T$ by a similar argument.