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Let $V$ be a finite dimensional vector space and $A$ and $B$ two linear transformations on $V$ such that $A^{2}=B^{2}=0$ and $AB+BA=1$.

1) Prove that if $N_{A}$ and $N_{B}$ are respective null spaces of $A$ and $B$, then $N_{A}=AN_{B}$, $N_{B}=BN_{A}$, and $V=N_{A}\oplus N_{B}$.

2) Prove the dimension of $V$ is even.

3) Prove that if the dimension of $V$ is 2, then $V$ has a basis with respect to which $A$ and $B$ are represented by matrices $(0,1),(0,0)$ and $(0,0),(1,0)$ (sorry I do not know how to type matrices).

My main difficulty is to prove $N_{A}\oplus N_{B}=V$. It follows trivially that $N_{A}\cap N_{B}=\{0\}$, but I do not know how to prove $N_{A}\oplus N_{B}=V$, and I also do not know how to prove $n=2k$.

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    You can use (1) to prove (2). Note that the transformation $A|_{N_B} : N_B \to N_A$ gives you an isomorphism (you have proven surjectivity, injectivity follows from $N_A \cap N_B = \{0\}$). The claim follows.2012-07-28

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$AB+BA=I$ $A(AB+BA)=A$ $ABA=A$ $rank(ABA)=rank(A)$ Combining with $rank(ABA)\leq \min(rank(AB),rank(A))$ Gives that $rank(A)\leq \min(rank(AB),rank(A))$

If $rank(AB), then $rank(A)$ is $\leq$ something strictly smaller than itself, a contradiction. So $rank(A)\leq rank(AB)$.

$rank(A)\leq rank(AB)\leq \min(rank(A),rank(B))$

Reusing the previous argument, we find that $rank(A)\leq rank(B)$.

We can start back up at the top with $AB+BA=I$ $B(AB+BA)=B$ $BAB=B$

and carry through the whole previous argument again (swapping $B$ and $A$ essentially) to find that $rank(B)\leq rank(A)$.

Both of these together imply that $rank(A)=rank(B)$. Since the nullspaces span the space together, the space must be of even dimension.