# Linear transformations $f, g:V\to V$ with properties $f \circ f = g \circ g = 0_V$ and $f \circ g + g \circ f = 1_V$ imply $\dim V$ is even?

Let's say you have $$V$$, a vector space. It is $$n$$-dimensional. Also, linear transformations $$f,g : V \to V$$ are such that $$f \circ f = g \circ g = 0_V$$ and $$f \circ g + g \circ f = 1_V$$.

How do you prove that $$\dim V$$ is even?

Ok, the real question is what $$0_V$$ and $$1_V$$ are? Null vector and the vector $$\begin{pmatrix}1\\ \vdots\\1\\\end{pmatrix}$$ ?

Now how do you prove it?

EDIT_2:
One more thing:
Suppose $$dim_\mathbb k V = 2$$, therefore exists a base $$\mathcal B$$ in $$V$$ such that:
$$M(f)_\mathcal B = \begin{pmatrix}0 & 0\\1 & 0\\\end{pmatrix}$$ , $$M(g)_\mathcal B = \begin{pmatrix}0 & 1\\0 & 0\\\end{pmatrix}$$

• $0_V$ is the zero map $\vec v\mapsto \vec 0$ of the vector space and $1_V$ is the identity map $\vec v \mapsto \vec v$. – lulu Nov 9 '18 at 17:34
• $0_V$ is the $0$ function on $V$ (ie. the function where $v\mapsto 0$ for all $v$). $1_V$ is the identity function on $V$ (ie, $v\mapsto v$) – memerson Nov 9 '18 at 17:35

Here is a sketch of my proof of the statement. Note that I use the notation $$\text{id}_V$$ instead of $$1_V$$.

We first observe that $$\text{ker}(f)\cap\ker(g)=0$$. Let $$x\in \text{ker}(f)\cap\ker(g)$$. By the condition $$f\circ g+g\circ f=\text{id}_V$$, we have $$x=(f\circ g+g\circ f)(x)=f\big(g(x)\big)+g\big(f(x)\big)=f(0)+g(0)=0.$$ Since $$\text{im}(f)\subseteq \ker(f)$$ and $$\text{im}(g)\subseteq \ker(g)$$ (due to the conditions $$f\circ f=0_V$$ and $$g\circ g=0_V$$), we obtain $$V=\text{im}(\text{id}_V)=\text{im}(f\circ g+g\circ f)\subseteq \text{im}(f)+\text{im}(g)\subseteq \ker(f)+\ker(g)\,.$$ Ergo, $$\ker(f)+\ker(g)=V$$ and $$\ker(f)\cap\ker(g)=0$$. Consequently, $$\ker(f)\oplus\ker(g)=V$$, as well as $$\text{im}(f)=\ker(f)$$ and $$\text{im}(g)=\ker(g)$$.

By the First Isomorphism Theorem for Vector Spaces, we have that $$\tilde{f}:=f|_{\operatorname{im}(g)}$$ is an isomorphism from $$\text{im}(g)=\ker(g)$$ to $$\text{im}(f)=\ker(f)$$. Similarly, $$\tilde{g}:=g|_{\text{im}(f)}$$ is an isomorphism from $$\text{im}(f)=\ker(f)$$ to $$\text{im}(g)=\ker(g)$$. Indeed, $$\tilde{g}=\tilde{f}^{-1}$$. In particular, if $$V$$ is finite-dimensional, then $$f$$ and $$g$$ have equal rank, and so $$\dim(V)=\text{rank}(f)+\text{rank}(g)$$ is an even integer.

$$0_V$$ is the function $$V \to V$$ such that $$0_V(v) = 0$$ for all $$v \in V$$ (where $$0$$ is the zero vector in $$V$$).

$$1_V$$ is the function $$V \to V$$ such that $$1_V(v) = v$$ for all $$v \in V$$.

Both of these are linear operators.

Hint Since $$f^2 = 0$$, $$\operatorname{im} f \subseteq \ker f$$, so $$\operatorname{rank} f \leq \dim\ker f$$, and the Rank-Nullity Theorem then implies that $$\operatorname{rank} f \leq \frac{1}{2} n$$. By symmetry $$\operatorname{rank} g \leq \frac{1}{2}n$$, too.

Now, $$n = \operatorname{rank} 1_V = \operatorname{rank}(fg + gf) \leq \operatorname{rank}(fg) + \operatorname{rank}(gf) .$$ We have $$\operatorname{rank}(fg) \leq \min(\operatorname{rank} f, \operatorname{rank} g) \leq \frac{1}{2} n$$, and by symmetry $$\operatorname{rank}(gf) \leq \frac{1}{2} n$$, so $$\operatorname{rank}(fg) + \operatorname{rank}(gf) \leq n$$, which forces all of the inequalities to be equalities. In particular, $$n$$ is even.

Here:

• $$0_V$$ denotes the zero transformation $$V \to V$$, $$v \mapsto 0$$, and
• $$1_V$$ denotes the identity transformation $$V \to V$$, $$v \mapsto v$$.

Hint Since $$f^2 = 0_V$$, there is a basis for which $$f$$ has Jordan normal form matrix representation $$[f] = \pmatrix{0_{k \times k}\\&\pmatrix{0&1\\&0}^{\oplus \ell}} ,$$ where $$k + 2 \ell = n$$ . Now, write $$[g]$$ in block matrix notation, with block sizes $$k \times k$$ and $$(2 \ell) \times (2 \ell)$$.

Expanding gives $$[1_V] = [fg + gf] = [f][g] + [g][f] = \pmatrix{0_{k \times k} & \ast\\ \ast&\ast} .$$ Since $$[1_V]$$ is the identity matrix, $$k = 0$$, so $$n = 2 \ell$$, and in particular $$n$$ is even. (Remark This proof shows that the hypothesis $$g^2 = 0_V$$ is unnecessary.)

• How do you know you the nullify of $g$ is at least $k$. And do you know you can block diagonalize $f$ and $g$ simultaneously? – Callus - Reinstate Monica Nov 9 '18 at 23:50
• The proof does not require knowing the nullity of $g$, nor does it use that $g^2 = 0$ for that matter. The proof also does not use simultaneous block-diagonalization of $f$ and $g$ (though it turns out that the hypotheses do imply that that is possible). – Travis Willse Nov 10 '18 at 0:10
• Oh I see. I thought you were suggesting to write $g$ in Jordan blocks, but you just said in blocks. This is nice, and illuminating. Thanks. – Callus - Reinstate Monica Nov 10 '18 at 2:11
• Thanks, and you're welcome. – Travis Willse Nov 10 '18 at 3:09