A question about super vector space 
Let $V=V_0\oplus V_1$ be a superspace, where its even subspace $V_0$
  and odd subspace $V_1$ both are $k$-dimensional. Let
  $(e_1,\cdots,e_k|f_1,\cdots,f_k)$ be a basis of $V$.
Consider linear transformations $\varphi,\psi$ on $V$, such
  that $\varphi(V_0)\subseteq V_0, \varphi(V_1)\subseteq V_1$;
  $\psi(V_0)\subseteq V_1,\psi(V_1)\subseteq V_0$.
Suppose $\varphi$ is nilpotent and $\varphi\psi-\psi\varphi=\theta$, where $\theta(e_i)=f_i$ and $\theta(f_i)=-e_i$.
  Then is it true that ${rank}(\varphi_0)={rank}(\varphi_1)$? Here
  $\varphi_0$ denotes restricting on even subspace , i.e.
  $\varphi|_{V_0}$.

If $\varphi^2=0$, then we know $\varphi$ anti-commute with $\theta$. Then $\varphi_0\theta_1=-\theta_0\varphi_1$, clearly ${rank}(\varphi_0)={rank}(\varphi_1)$. Is this true in general, when $\varphi^n=0$ with arbitrary $n$?
Thanks!
 A: $\DeclareMathOperator{\rank}{rank} \require{cancel}$
Let $a \in \ker(\varphi_0)$. Then
$$[\varphi, \psi](a, 0) = \varphi\circ\psi(a, 0) - \cancelto{(0, 0)}{\psi\circ\varphi(a, 0)} = \theta(a, 0)$$
Hence
$$\varphi\circ\psi(a, 0) = \theta(a, 0) = (0, a)$$
Since $\varphi$ is $n$-nilpotent
$$\varphi^{n-1}(0, a) = \varphi^{n-1}(\varphi\circ\psi(a, 0)) = \varphi^{n}\circ\psi(a, 0) = (0, 0)$$
This means that $a \in \ker(\varphi^{n-1}_1)$ which implies
$$\ker(\varphi_0) \subseteq \ker(\varphi^{n-1}_1)$$
By the analogous reasoning we are also able to conclude that
$$\ker(\varphi_1) \subseteq \ker(\varphi^{n-1}_0)$$
As you noted, when $n=2$ it is immediate to see that $\ker(\varphi_0) = \ker(\varphi_1)$ whence follow the thesis. However this may not be the case for $n = 3$. Consider for example $V_0 = V_1 = \mathbb{R}^3_{\mathbb{R}}$ with its canonical basis and the linear transformation represented by the upper triangular matrix
$$T = \begin{bmatrix}
 0 & 1 & 1 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{bmatrix}
$$
Next choose $\varphi = T \oplus T^2$ and any $\psi$ such that $[\varphi, \psi] = \theta$, for instance
$$\psi : V_0 \oplus V_1 \to V_0 \oplus V_1 : (x, y) \mapsto (Ay, Bx)$$
where
$$A = \begin{bmatrix}
 -24 & -18 & 22 \\
 -19 & 6 & -6 \\
 0 & 0 & -48 \\
\end{bmatrix}
\qquad%
B = \begin{bmatrix}
 -42 & -12 & -28 \\
 0 & -5 & -23 \\
 0 & -19 & -13 \\
\end{bmatrix}$$
Thus all hypotheses are met but
$$\dim\ker(\varphi_0) = 1 < 2 = \dim\ker(\varphi_1)$$
Therefore
$$\rank(\varphi_1) = 1 \neq 2 = \rank(\varphi_0)$$
