Alternating Bilinear form, nilpotent map problem: Where is the error?

Question

The problem is the following:

Let $V$ be a finite dimensional $\mathbb{C}$-vector space, and $b : V \times V \to \mathbb{C}$ a non- degenerate symmetric bilinear form. Let $f : V \to V$ be a nilpotent $\mathbb{C}$-linear map such that $$ b(f(x), y) = -b(x, f(y)) $$ for all $x, y$ in $V$ . If $ \dim \ker f = 1$ show that $\dim V$ is odd.

My (slightly flawed) solution attempt is the following: Define $b_f(x,y) = b(f(x),y)$. Then, the fact that $ b(f(x), y) = -b(x, f(y)) $ implies that $b_f$ is a skew-symmetric bilinear form. Let $W$ be any vector space complement to $\ker(f)$. Suppose that $b_f(x,y) =0$, for all $y \in V$. Then, $b(f(x), y) =0 $ for all $y \in V$, so that $f(x) =0$. Now, as I was typing I realized here was my error because I did not have the quantifier on $y$ previously, so at this point I concluded that $b_f \mid_{W \times W}$ is non-degenerate, and so $W$ is even dimensional and then I would be done.

Since this proof is wrong:

1. Is this even on the right track?

2. What is a correct proof, or hint toward one?

Answer 1

You proved that : if $x\in W$ and $b_f(x,y) = 0$ for all $y\in V$, then $x=0$ (indeed, $f(x)=0$, so $x\in \ker f \cap W = 0$).

Now suppose the same holds only for $y\in W$; and let $z\in V$. Write $z=y+z'$ with $y\in W, z'\in \ker f$. What can you say about $b_f(x,z) $ ?

It follows that $b_{f\mid W\times W}$ is non-degenerate.

This amounts essentially to Omnomnomnom's comment, where they consider $V/\ker f$ instead of $W$ (in linear algebra over fields, quotients do essentially the same thing as complements)

Answer 2

Here is matrix proof for dimensions 2 to 4, that can help IMHO to understand the constraints of this issue. I assume that we are on $\mathbb{R}$.

1) It is not possible for dimension 2 :

Indeed, with respect to a suitable basis, one can assume that the nilpotent operator with dim ker f = 1 is represented by the following matrix :

$$F:=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$

With respect to this basis, let :

$$B:=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

be the (symmetric) matrix of the bilinear form $b$.

Let us write now transcript condition $b(f(X), Y) = -b(X, f(Y))$ under matrix form.

For any $X=(x_1,y_1)^T$ and any $Y=(x_2,y_2)^T$, we must have :

$$\text{for all} X,Y, \ \ \ \ (FX)^TBY=-X^TB(FY)\tag{1}$$

i.e.,

$$\text{for all} X,Y, \ \ \ \ X^T(F^TB)Y=-X^T(BF)Y\tag{2}$$

Condition (2) can be written under the simplified form :

$$F^TB=-BF\tag{3}$$

Giving

$$\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}=-\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\tag{4}$$

Identifying the matrices' entries we find $b=c=0$, giving the following matrix for the bilinear form :

$$B:=\begin{pmatrix}a&b\\b&c\end{pmatrix}=\begin{pmatrix}a&0\\0&0\end{pmatrix}$$

which is degenerate, in contradiction with assumptions.

2) It is possible in dimension 3 :

The equivalent of matrix $F$ above is now :

$$F=\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}$$

giving the following condition

$$\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}=-\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\tag{5}$$

All computations done, we find $a=b=e=f=0$ and $d=-c$, giving a non-degenerate solution (assuming $c \neq 0$ of course):

$$B=\begin{pmatrix}0&0&-c\\0&c&0\\-c&0&0\end{pmatrix}$$

Please note that there is not much room for choice : up to a factor, there is a unique such form.

Remark : About matrices of nilpotent linear maps, see for example the answer in How to prove that a nilpotent operator has a basis representation that is strictly upper triangular? .

4) It is not possible for dimension 4 :

Because :

$$\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}\begin{pmatrix}a&b&d&g\\b&c&e&h\\d&e&f&i\\g&h&i&j\end{pmatrix}=-\begin{pmatrix}a&b&d&g\\b&c&e&h\\d&e&f&i\\g&h&i&j\end{pmatrix}\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}\tag{6}$$

gives $a=b=c=d=e=g=0$, i.e., the matrix of the bilinear form would be degenerated, once again for the even case.

Remarks :

1) It is not uninteresting to remark that (3), written under the equivalent form:

$$BFB^{-1}=-F^T$$

implies that $\det(F-xI)=\det(-F^T-xI)=(-1)^n \det((F+xI)^T)=(-1)^n \det((F+xI))$. Therefore, if we denote by $P(x)$ the characteristic polynomial of $F$, we have $P(x)=P(-x)$ in the even case, and $P(x)=-P(-x)$ in the odd case in conformity with the fact that a nilpotent operator has characteristic polynomial $(-1)^nx^n$.

2) We have considered nilpotent linear maps of order $n-1$ (the annihilation takes place at the $(n-1)$th iteration). For a more general case, one should consider strictly upper triangular matrices.

3) Proving the possibility or impossibility of the existence (according to the parity of the ambient space) of the bilinear form in the general case looks feasible but I don't see a way to do it 'cleanly' (maybe by matrices' partitionning into blocks or by considering (3) as a Sylvester equation of some kind ?).