# Asking a linear algebra problem in Berkeley Problems in Mathematics, Spring 1986.

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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Since $(AB+BA)x = x$, let $x_a = ABx$, $x_b = BAx$, then $x= x_a + x_b$, and $A x_a = 0$, $B x_b$ = 0. Since $\ker A \cap \ker B = \{0 \}$ (also follows from $AB+BA = I$), this shows that $V = \ker A \oplus \ker B$. – copper.hat Jul 28 '12 at 19:18
Thanks you! Now I can try to solve 2). – Bombyx mori Jul 28 '12 at 19:20
@user32240 You can type matrices as follows: \begin{pmatrix} a & b\\ c & d \end{pmatrix} gives $\begin{pmatrix} a & b\\ c & d \end{pmatrix}$. – martini Jul 28 '12 at 19:31
thx!==========! – Bombyx mori Jul 28 '12 at 19:35
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. – Jim Jul 28 '12 at 19:58

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

If $rank(AB)<rank(A)$, 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.

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