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 $\begin{pmatrix}1\\ \vdots\\1\\\end{pmatrix}$ ?

EDIT: Thanks for the answers!
Now how do you prove it?

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

  • 1
    $\begingroup$ $0_V$ is the zero map $\vec v\mapsto \vec 0$ of the vector space and $1_V$ is the identity map $\vec v \mapsto \vec v$. $\endgroup$ – lulu Nov 9 '18 at 17:34
  • 1
    $\begingroup$ $0_V$ 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, $v\mapsto v$) $\endgroup$ – 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.



  • $0_V$ 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.)

  • $\begingroup$ 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? $\endgroup$ – Callus - Reinstate Monica Nov 9 '18 at 23:50
  • $\begingroup$ 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). $\endgroup$ – Travis Willse Nov 10 '18 at 0:10
  • $\begingroup$ 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. $\endgroup$ – Callus - Reinstate Monica Nov 10 '18 at 2:11
  • $\begingroup$ Thanks, and you're welcome. $\endgroup$ – Travis Willse Nov 10 '18 at 3:09

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