Space of invariant bilinear forms

Abstract

Let ${\mathbb {F}}$ be a field, V a vector space of dimension n over ${\mathbb {F}}$. Then the set of bilinear forms on V forms a vector space of dimension $n^2$ over ${\mathbb {F}}$. For char ${\mathbb {F}}\ne 2$, if T is an invertible linear map from V onto V then the set of T-invariant bilinear forms, forms a subspace of this space of forms. In this paper, we compute the dimension of T-invariant bilinear forms over ${\mathbb {F}}$. Also we investigate similar type of questions for the infinitesimally T-invariant bilinear forms (T-skew symmetric forms). Moreover, we discuss the existence of nondegenerate invariant (resp. infinitesimally invariant) bilinear forms.

1 Introduction

Let ${\mathbb {F}}$ be a field of characteristic $\not =2$, $n\ge 1$ an integer. Let V be a vector space of dimension n over ${\mathbb {F}}$. Then the set of bilinear forms on V forms a vector space of dimension $n^2$ over ${\mathbb {F}}$. If T is an invertible linear map from V onto V then the set ${\mathcal {B}}$ of T-invariant bilinear forms, forms a subspace of this space of forms. In this paper, we investigate the following question.

Question

What is the dimension of ${\mathcal {B}}$? Does there exist any non-degenerate form in ${\mathcal {B}}$?

Let V be a vector space over ${\mathbb {F}}$, then a bilinear form B on V is said to be invariant (resp. infinitesimally invariant) under a linear map $T:V\rightarrow V$, if for all $\displaystyle u, v\in V$, $B(Tu,Tv)=B(u,v)$ (resp. $B(Tu,v)+B(u,Tv)=0$). A bilinear form B on V is said to be symmetric, if for all $\displaystyle u, v\in V$, $B(u,v)=B(v,u)$. If $\text{ char }\, {\mathbb {F}}\ne 2$, then a bilinear form B on V is said to be skew-symmetric, if for all $\displaystyle u,v\in V$, $B(u,v)=-B(v,u)$.

The starting point in this study is the existence of such T-invariant forms. This problem has been answered in many ways in the literature. Sergeichuk [6] studied systems of forms and linear mappings by associating with them self-adjoint representations of a category with involution. Gongopadhyay and Kulkarni [2] investigated the question of existence of T-invariant non-degenerate symmetric (resp. skew-symmetric) bilinear forms. They also answered the infinitesimal version of this question. In the literature, the problem to find necessary and sufficient conditions for an endomorphism to be self-adjoint, skew-selfadjoint or orthogonal for at least one non-degenerate quadratic form (or for a symplectic form) has also acquired an important place. Frobenius [1] proved that every endomorphism of a finite dimensional vector space V is self-adjoint for at least one non-degenerate symmetric bilinear form on V. Later, Stenzel [7] determined when an endomorphism could be skew-selfadjoint for a non-degenerate quadratic form, or self-adjoint or skew-selfadjoint for a symplectic form on complex vector spaces. However his results were later generalized to an arbitrary field [3]. Pazzis [5] tackled the case of the automorphisms of a finite dimensional vector space that are orthogonal (resp. symplectic) for at least one non-degenerate quadratic form (resp. symplectic form) over an arbitrary field of characteristics 2.

In this paper, we investigate the question of computing the dimension of the space of bilinear forms invariant (resp. infinitesimal invariant) under a given invertible linear transformation (resp. linear transformation) on a finite dimensional vector space over a field of characteristic $\not =2$. We have answered this question for several possible types of linear transformations and also discussed about the existence of non-degenerate forms in every case.

2 Preliminaries

Let $f(x)\in {\mathbb {F}}[x]$, f(x) monic, $\deg f=n$ and $\displaystyle 0, \pm 1$ are not the roots. Then $f^*(x)=x^nf(x^{-1})$ is called as the reciprocal of f (see also [2]). A monic polynomial $f(x)\in {\mathbb {F}}[x]$, $f(0)\ne 0\ne f(\pm 1)$ is said to be self-reciprocal if $f(x)=f^*(x)$.

Lemma 2.1

Let $f(x)\in {\mathbb {F}}[x]$ be monic, $f(0)\ne 0\ne f(\pm 1)$. If $\alpha \in {\bar{{\mathbb {F}}}}$ is a root of f(x) of multiplicity m, then $\alpha ^{-1}$ is a root of $f^*(x)$ of the same multiplicity, i.e., m.

Proof

Easy to see. $\square$

Lemma 2.2

Let $f(x)\in {\mathbb {F}}[x]$ be self-reciprocal, $\deg f=n$, then n is even.

Proof

Clear. $\square$

If $p(x)=\sum _{i=0}^{2n}c_ix^i\in {\mathbb {F}}[x]$ is a self-reciprocal polynomial, then $c_0=c_{2n}=1$ and for $1\le i\le n$, $c_i=c_{2n-i}$.

DEFINITION 2.3

A linear operator $T:V\rightarrow V$ is said to be self-reciprocal if its characteristic polynomial $\chi _T(x)$ is self-reciprocal.

It is important to observe that self-reciprocal operators exist only on even dimensional vector spaces.

Now for $T\in \mathrm{End}(V)$, let us study bilinear form B on V satisfying

$$\begin{aligned} B(Tv,w)+B(v,Tw)=0 \,\,\quad \text{ for } \text{ all } v,w\in V \,\text{(i.e. }\, T\text{-skew } \text{ symmetric } \text{ forms) }. \end{aligned}$$

Let $f(x)\in {\mathbb {F}}[x]$, f(x) monic, deg $f=n$ and 0 is not a root of f, then $f^-(x)=(-1)^nf(-x)$ is called as the additive-dual of f. Let $f(x)=x^{\epsilon }f_0(x)$, $\epsilon \ge 0$ such that $f_0(0)\ne 0$ and $f_0$ monic. We call $f_0$ the reduced part of f. A monic polynomial $f(x)\in {\mathbb {F}}[x]$, $f(0)\ne 0$ is said to be additive self-dual ($\alpha$-self dual for short) if $f(x)=f^-(x)$.

DEFINITION 2.4

A linear operator $T:V\rightarrow V$ is said to be $\alpha$-self dual if its characteristic polynomial is so.

Note that all the results in the invariant case have their infinitesimal counterparts. In the last section of this paper we only state the main results in the infinitesimal set up.

2.1 Dimension invariance over field extensions

Let V be a vector space over ${\mathbb {F}}$ of dimension n, $T\in \text {End}(V)$. The characteristic polynomial $\chi _T(x)\in {\mathbb {F}}[x]$ of T can be expressed uniquely (up to permutations of the factors) as $\displaystyle \chi _T(x)=\Pi _{i=1}^rf_i(x),\,\,f_i(x)\mid f_{i-1}(x)$ and $T|_{V_{f_i(x)}=\{v\in V|f_i(T)v=0\}}$ is cyclic for $2\le i\le r$. The polynomials $f_i(x)$ are the invariant factors of T and this factorization remains invariant under field extensions. There is an ordered basis $\displaystyle (e_1,\ldots , e_n)$ of V over ${\mathbb {F}}$ with respect to which the matrix representation of T is given as

$$\begin{aligned} R=\left( \begin{array}{llll} C(f_1(x)) &{}&{}&{}\\ &{}\ddots &{}&{}\\ &{}&{}\ddots &{}\\ &{}&{}&{}C(f_r(x)) \end{array}\right) , \end{aligned}$$

where for $1\le i\le r$, $C(f_i(x))$ is a companion matrix of $f_i(x)$ and R is the rational canonical form of T. Under field extensions this form remains the same. In the next proposition, we show that the dimension of T-invariant (resp. infinitesimally invariant) bilinear forms remains invariant under field extensions of ${\mathbb {F}}$.

PROPOSITION 2.5

Let V be a n dimensional vector space over ${\mathbb {F}},$ $T\in \mathrm{End}(V)$. Then dimension of the space of T-invariant (resp. infinitesimally invariant) bilinear forms remains invariant under field extensions of ${\mathbb {F}}$.

Proof

Without loss of generality, take $V={\mathbb {F}}^n$. Then for $\mathbb K\supseteq {\mathbb {F}}$ a field extension, $V'={\mathbb {K}}^n$ is its extension as a ${\mathbb {K}}$ vector space. Let $\displaystyle (e_1,\ldots , e_n)$ be the ordered basis of V over ${\mathbb {F}}$ with respect to which T has the rational canonical form R as expressed above. This will still be an ordered basis of $V'$ over ${\mathbb {K}}$. Let $T'\in \text {End}(V')$ be determined by R. Let ${\mathcal {B}}=\{B \mid B(T'u',T'v')=B(u',v')$ (resp. $B(T'u',v')+B(u',T'v')=0)$ $\forall \,\,u',v'\in V'\}$ be the space of $T'$-invariant (resp. infinitesimally $T'$-invariant) forms over ${\mathbb {K}}$. Then ${\mathcal {B}}=\{X=(B(e_i,e_j)=x_{ij})\in M_{n}({\mathbb {K}})\mid R^tXR=X$ (resp. $R^tX+XR=0)\}$, which is the solution space of a system of $n^2$ linear homogeneous equations having coefficients in ${\mathbb {F}}$ (as $R\in M_{n}({\mathbb {F}})$), so is a vector space over ${\mathbb {F}}$ with $\dim \mathcal B|_{{\mathbb {K}}}=\dim {\mathcal {B}}|_{{\mathbb {F}}}$. $\square$

3 Invariance under indecomposable operators

Now let us discuss bilinear forms invariant under some indecomposable transformations. We consider two cases: (1) T or $-T$ unipotent and (2) T and $-T$ nonunipotent.

Case 1. T or $-T$ unipotent. Since invariant forms of T or $-T$ are the same, we may assume T as unipotent. Let $V={{\mathbb {F}}}^n$, $e_1,\ldots ,e_n\in V$ be the standard bases elements. Let $T:V\rightarrow V$ be a linear transformation with the Jordan block as

$$\begin{aligned} C=\left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l@{\quad }l} 1&{}0&{}0&{}\cdots &{}0\\ 1&{}1&{}0&{}\cdots &{}0\\ 0&{}1&{}1&{}\cdots &{}0\\ \vdots &{} &{}\quad \ddots &{}\ddots &{}\\ 0&{}0&{}0&{}\cdots &{}1 \end{array}\right) . \end{aligned}$$

Then we have $\displaystyle Te_n=e_n$ and $\displaystyle \forall \, i,\, 1\le i\le n-1,\,\, Te_i=e_i+e_{i+1}\,$. Now let $B:V\times V\rightarrow k$ be any bilinear form invariant under T. Again, we have the equation

$$\begin{aligned} \left( B(e_i,e_j)\right) =X=C^tXC=\left( B(Ce_i,Ce_j)\right) =\left( B(T e_i,T e_j)\right) . \end{aligned}$$

Here $C=I+N$, thus the above equation becomes $(I+N^t)X(I+N)=X$ which implies and is implied by $N^tX+XN+N^tXN=0$. Similarly for $C=-I+N$, we have $-N^tX-XN^t+N^tXN=0$.

Theorem 3.1

Let V be a vector space of dimension n over ${\mathbb {F}}$. If T is an indecomposable invertible transformation from V onto V such that T or $-T$ is unipotent, then the space of T-invariant bilinear forms is of dimension n. If n is even (resp. odd) then there exists a nondegenerate T-invariant skew symmetric (resp. symmetric) but not a symmetric (resp. skew-symmetric) bilinear form.

Proof

Let us consider C as above with the standard basis $\displaystyle \{e_1,\ldots ,e_n\}$ of V over ${\mathbb {F}}$ and B be a T -invariant bilinear form with $(B(e_i, e_j))=X$.

Case i. $i=n$: $x_{n,j}=B(e_n,e_j)=B(Ce_n,Ce_j)$ for $1\le j\le n$. Thus we have $x_{n,j}=x_{n,j}+x_{n,j+1}$ for $1\le j\le n-1$, i.e., $x_{n,j+1}=0$ for $1\le j\le n-1$.

Case ii. $j=n$: $x_{i,n}=B(e_i,e_n)=B(Ce_i,Ce_n)$ for $1\le i\le n$. Thus we have $x_{i,n}=x_{i,n}+x_{i+1,n}$ for $1\le i\le n-1$, i.e., $x_{i+1,n}=0$ for $1\le i\le n-1$.

Case iii. $1\le i,j\le n-1$: $x_{i,j}=B(e_i,e_j)=B(Ce_i,Ce_j)=B(e_i+e_{i+1},e_j+e_{j+1})$ i.e., $x_{i,j}=x_{i,j}+x_{i,j+1}+x_{i+1,j}+x_{i+1,j+1}$, which says that $x_{i,j+1}=-x_{i+1,j}-x_{i+1,j+1}$ for $1\le i,j\le n-1$.

Thus recursively we have $x_{n-i,j+1}=0$ for $i+1\le j\le n-1$ and $x_{n-(s-1),s}=(-1)^{s-1}x_{n,1}$ for $1\le s\le n-1$. So,

$$\begin{aligned} X=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} x_{1,1}&{}x_{1,2}&{}x_{1,3}&{}\cdots &{}x_{1,n-1}&{}(-1)^{n-1}x_{n,1}\\ x_{2,1}&{}x_{2,2}&{}x_{2,3}&{}\cdots &{}(-1)^{n-2}x_{n,1}&{}0\\ \vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots &{}\vdots \\ x_{n-1,1}&{}(-1)x_{n,1}&{}0&{}\cdots &{}0&{}0\\ x_{n,1}&{}0&{}0&{}\cdots &{}0&{}0 \end{array}\right) . \end{aligned}$$

Note that $\det X=x_{n,1}^n$, B is non degenerate when $x_{n,1}$ is nonzero.

Further, we have $x_{n-2,2}=(-1)x_{n-1,1}+(-1)^2x_{n,1}$, $x_{n-3,3}=(-1)^2x_{n-1,1}+2(-1)^3x_{n,1}$ and thus recursively we get for $1\le t\le n-1$, $x_{n-t,t}=(-1)^{t-1}x_{n-1,1}+(t-1)(-1)^tx_{n,1}$, so

$$\begin{aligned} x_{1,n-1}=(-1)^{n-2}x_{n-1,1}+(n-2)(-1)^{n-1}x_{n,1}. \end{aligned}$$

(3.1)

Similarly $x_{n-3,2}=-x_{n-2,1}-x_{n-2,2}=(-1)x_{n-2,1}+(-1)^2x_{n-1,1}+(-1)^3x_{n,1}$, $x_{n-4,3}=-x_{n-3,2}-x_{n-3,3}=(-1)^2x_{n-2,1}+2(-1)^3x_{n-1,1}+3(-1)^4x_{n,1}$ and thus recursively for $2\le t\le n-1$, we get $x_{n-t,t-1}=(-1)^{t-2}x_{n-2,1}+(t-2)(-1)^{t-1}x_{n-1,1}+(t-1)(-1)^tx_{n,1}$, so

$$\begin{aligned} x_{1,n-2}=(-1)^{n-3}x_{n-2,1}+(n-3)(-1)^{n-2}x_{n-1,1}+(n-2)(-1)^{n-1}x_{n,1}.\nonumber \\ \end{aligned}$$

(3.2)

Repeating this process recursively, we get for every $1\le s\le n-2$ and $s\le t\le n-1$,

$$\begin{aligned} x_{n-t,t-s+1}=(-1)^{t-s}x_{n-s,1}+\sum _{i=0}^{s-1}(t-s+i)(-1)^{t-s+i+1}x_{n-s+i+1,1}, \end{aligned}$$

(3.3)

i.e., for $1\le s\le n-2$, we have

$$\begin{aligned} x_{1,n-s}= & {} (-1)^{n-s-1}x_{n-s,1}+ \sum _{i=0}^{s-1}(n-s-1+i)(-1)^{n-s+i}x_{n-s+1+i,1}. \end{aligned}$$

(3.4)

Thus basis of the space of invariant forms over ${\mathbb {F}}$ is

$$\begin{aligned}&\left\{ e_{n,1}+\sum _{s=1}^{n-2}\sum _{t=s}^{n-1}(t-1)(-1)^te_{n-t,t-s+1}, e_{n-1,1}+\sum _{t=2}^{n-1}(-1)^{t-1}e_{n-t,t}\right. \\&\quad +\displaystyle \sum _{s=2}^{n-2}\sum _{t=s}^{n-1}(t-2)(-1)^{t-1}e_{n-t,t-s+1},\ldots , e_{n-l,1}+\sum _{t=l+1}^{n-1}(-1)^{t-l}e_{n-t, t-l+1}\\&\quad \left. +\sum _{s=l+1}^{n-2}\sum _{t=s}^{n-1}(t-l-1)(-1)^{t-l}e_{n-t,t-s+1},\ldots , e_{2,1}-e_{1,2}, e_{1,1}\right\} \end{aligned}$$

and X has a unique expression as a linear combination of these vectors with $\displaystyle x_{n,1}, x_{n,2}, \ldots , x_{2,1}, x_{1,1}$ respectively as free coefficients.

Note that if n is even then B symmetric (i.e., $X^t=X$) implies $x_{n,1}=0$, i.e., B cannot be nondegenerate symmetric form. However we have plenty of nondegenerate T-invariant skew symmetric forms obtainable by taking $x_{n,1}\not =0,\,x_{i,j}=-x_{j,i}$ and substituting in X.

Similarly if n is odd, then B skew symmetric (i.e. $X^t=-X$) implies $x_{n,1}=0$, so B cannot be a nondegenerate skew symmetric form. However we have plenty of nondegenerate T-invariant symmetric forms obtainable by taking $x_{n,1}\not =0,\,x_{i,j}=x_{j,i}$ and substituting in X. $\square$

COROLLARY 3.2

With the hypothesis as in the theorem,  if n is odd,  then the subspace of   T-invariant symmetric forms is of dimension $\frac{n+1}{2}$ and the subspace of T-invariant skew symmetric forms is of dimension $\frac{n-1}{2}$.

Proof

Let $X=(x_{i,j})$ be a T-invariant bilinear form. If X is symmetric, then as n is odd from equation (3.4), we get

$$\begin{aligned} x_{n-s,1}= & {} x_{1,n-s}=(-1)^sx_{n-s,1}+(n-s-1)(-1)^{s-1}x_{n-s+1,1}\\&\quad +\cdots +(n-2)x_{n,1}. \end{aligned}$$

If s is odd, i.e., $n-s$ is even, then

$$\begin{aligned} x_{n-s,1}=\frac{1}{2}[(n-s-1)(-1)^{s-1}x_{n-s+1,1}+\cdots +(n-2)x_{n,1}], \end{aligned}$$

however there is no condition on $x_{n-s,1}$ for $n-s$ odd. Thus with similar steps as in the theorem one may calculate the basis with $\displaystyle \{x_{1,1},x_{3,1},\ldots ,x_{n,1}\}$ respectively as the free coefficients for the T-invariant symmetric forms. So subspace of T-invariant symmetric bilinear forms is of dimension $\frac{n+1}{2}$.

Similarly $\displaystyle \{x_{2,1},x_{4,1},\ldots ,x_{n-1,1}\}$ is the largest set of free coefficients for the T-invariant skew symmetric forms. So the subspace of T-invariant skew symmetric forms is of dimension $\frac{n-1}{2}$. $\square$

COROLLARY 3.3

With the hypothesis as in the theorem,  if n is even,  then the subspace of T-invariant symmetric forms is of dimension $\frac{n}{2}$ and the subspace of T-invariant skew symmetric forms is of dimension $\frac{n}{2}$.

Proof

Similar. $\square$

Case 2. T and $-T$ nonunipotent. In light of Proposition 2.5, for ${\mathbb {F}}$ algebraically closed (and hence a perfect field) we know (Theorem 15.3, J E Humphreys, Linear Alg. Groups, p. 99) that for $T\in O(V,B)$ (the orthogonal group O(V, B), [2] being an algebraic group), if $T=T_sU$ is the Jordan decomposition of T with $\displaystyle T_s,U$ semisimple and unipotent respectively, then $T_s, U\in O(V,B)$. In the next lemma, we use this fact to find the bilinear forms invariant under nonunipotent linear operators.

Lemma 3.4

Let V be a vector space of dimension n over ${\mathbb {F}}$. If   T is an indecomposable, non-unipotent invertible transformation from V onto V such that the characteristic polynomial $\chi _T(x)=(x-\alpha )^n$, $\pm 1\ne \alpha \in {\mathbb {F}}$, then the space of T-invariant bilinear forms ${\mathcal {B}}=0$.

Proof

Clear (see [2]). $\square$

4 Invariance under decomposable operators

Let us denote minimal polynomial of a linear operator T by $m_T(x)$.

Lemma 4.1

Let V be a finite dimensional vector space over ${\mathbb {F}}$, $T:V\rightarrow V$ an invertible linear operator, $\displaystyle \alpha ,\beta \in {\mathbb {F}}$, $\alpha \beta \ne 1$. If $(x-\alpha )^r$ and $(x-\beta )^s$ occur as elementary divisors of $\chi _T(x)$, then for B a T-invariant bilinear form on V for all $\displaystyle u\in V_{(x-\alpha )^r},\, v\in V_{(x-\beta )^s}$, we have $B(u,v)=0$.

Proof

Since $T\in O(V,B)$ implies $T_s\in O(V,B)$, so for all $\displaystyle u\in V_{(x-\alpha )^r},\, v\in V_{(x-\beta )^s}$, we have $B(T_su,T_sv)=B(u,v)$ or $(\alpha \beta -1)B(u,v)=0$, i.e., $B(u,v)=0.$ $\square$

Lemma 4.2

Let V be a finite dimensional vector space over ${\mathbb {F}}$, $T:V \rightarrow V$ an invertible linear operator and  $\pm 1\ne \alpha ,\,\,\beta \in {\mathbb {F}}$, $\alpha \beta =1$.  If  $(x-\alpha )^r$ and $(x-\beta )^s$ occur as elementary divisors of $\chi _T(x)=(x-\alpha )^r(x-\beta )^s=m_T(x)$, then dimension of the space of  T-invariant bilinear forms is $2\cdot \min \{r,s\}$.

Proof

Without loss of generality, we can assume that $V=V_{(x-\alpha )^r(x-\beta )^s}$. Then $\dim V=r+s=n$, $V=V_{(x-\alpha )^r}\oplus V_{(x-\beta )^s}$ and the Jordan form of T is

Let $\{e_1,\ldots , e_r, e_{r+1}, \ldots , e_{n=r+s}\}$ be the standard basis with $\{e_1,\ldots , e_r\}$ as a basis for $V_{(x-\alpha )^r}$ and $\{e_{r+1},\ldots , e_{r+s}\}$ as a basis for $V_{(x-\beta )^s}$. Then we have for $1\le i\le r-1,\,\,Te_i=\alpha e_i+e_{i+1}$, $Te_r=\alpha e_r$ and for $1\le j\le s-1,\,Te_{r+j}=\beta e_{r+j}+e_{r+j+1}$, $Te_n=\beta e_n$.

In light of Proposition 2.5, without loss of generality, we may assume that ${\mathbb {F}}$ is algebraically closed and hence a perfect field. Let $T=T_sU$ be the Jordan decomposition of T and B for any T-invariant bilinear form given by $X=(x_{ij})=(B(e_i,e_j))$. Then $T_s\in O(V,B)$, so for $v,w\in V_{(x-\alpha )^r}$, $B(v,w)=B(T_s v,T_sw)=\alpha ^2B(v,w)$, but as $\alpha \ne \pm 1$, $B(v,w)=0$. Thus $B|_{V_{(x-\alpha )^r}}=0$ and similarly one says that $B|_{V_{(x-\beta )^s}}=0$.

Now again, without loss of generality, one assumes that $r\le s$. Let us find the $r\times s$ block of X.

Case i. For $j=n$ and $1\le i\le r-1$, $B(Te_i,Te_n)=B(e_i,e_n)=x_{in}$ implies $x_{i+1n}=0$.

Case ii. For $i=r$ and $1\le j\le s-1$, $B(Te_r,Te_{r+j})=B(e_r,e_{r+j})=x_{rr+j}$ implies $x_{rr+j+1}=0$.

Case iii. For $1\le i\le r-1$ and $1\le j\le s-1$, $B(Te_i,Te_{r+j})=B(e_i,e_{r+j})=x_{ir+j}$ implies $\alpha x_{i r+j+1}+\beta x_{i+1 r+j}+x_{i+1 r+j+1}=0$, i.e.,

$$\begin{aligned} x_{ir+j+1}=-\frac{\beta }{\alpha }x_{i+1 r+j}-\frac{1}{\alpha }x_{i+1 r+j+1}. \end{aligned}$$

So $\{e_{ir+1}+\sum _{j=1,k>r+1}^{i-1}u_{jk}(\alpha ,\beta )e_{jk}\,:\,1\le i\le r\}$ forms a basis for this $r\times s$ block, where $e_{ij}$ is an elementary matrix of order n with 1 at ij-th entry and 0 elsewhere and $u_{ij}(\alpha , \beta )$ are some particular values in ${\mathbb {Q}}(\alpha , \beta )$.

Now let us find the $s\times r$ block of $X=(B(e_i,e_j))$.

Case i. For $i=n,\,\,1\le j\le r-1$, $B(Te_n,Te_j)=B(e_n,e_j)=x_{nj}$ implies $x_{n j+1}=0$.

Case ii. For $j=r,\,\,1\le i\le s-1$, $B(Te_{r+i},Te_r)=B(e_{r+i},e_r)=x_{r+i\, r}$ implies $x_{r+i\, r}+\alpha x_{r+i+1 \,r}=x_{r+i\, r}$, i.e., $x_{r+i+1\, r}=0$.

Case iii. For $1\le i\le s-1$ and $1\le j\le r-1$, $B(Te_{r+i},Te_j)=B(e_{r+i},e_j)=x_{r+i\,j}$ implies $\beta x_{r+i\,j+1}+\alpha x_{r+i+1\,j}+x_{r+i+1\,j+1}=0$, i.e.,

$$\begin{aligned} x_{r+i+1\,j}=-\frac{\beta }{\alpha }x_{r+i\,j+1}-\frac{1}{\alpha }x_{r+i+1\,j+1}. \end{aligned}$$

So $\{e_{r+1i}+\sum _{k>r+1,j=1}^{i-1}u_{kj}(\alpha ,\beta )e_{kj}\,:\,1\le i\le r\}$ forms a basis for this $s\times r$ block, where $u_{ij}(\alpha ,\beta )=u_{ji}(\alpha , \beta )$.

Thus dimension of space of T-invariant bilinear forms is $2r=2\cdot \min \{r,s\}$. $\square$

Note that the T-invariant bilinear forms obtained in the above lemma is always degenerate for $s\ne r$ as against, we have lots of nondegenerate forms for $s=r$ obtained in the paper.

COROLLARY 4.3

Under the hypothesis of the lemma,  if $(x-\alpha )^r$ and $(x-\beta )^s$ occur as elementary divisors of $\chi _T(x)=(x-\alpha )^r(x-\beta )^s,$ then the dimension of the subspace of T-invariant symmetric (resp. skew symmetric) bilinear forms is $\min \{r,s\}$.

Proof

Again, without loss of generality, we can assume that $V=V_{(x-\alpha )^r(x-\beta )^s}$ and denote the space of T-invariant bilinear forms by ${\mathcal {B}}=\{X=(x_{ij})\}$. Then by the lemma, $\{e_{ir+1}+e_{r+1i}+\sum _{j=1,k>r+1}^{j=i-1}u_{jk}(\alpha ,\beta )(e_{jk}+e_{kj})\,:\,1\le i\le r\}$ forms a basis for ${\mathcal {B}}$ if ${\mathcal {B}}$ is symmetric and $\{e_{ir+1}-e_{r+1i}+\sum _{j=1,k>r+1}^{j=i-1}u_{jk}(\alpha ,\beta )(e_{jk}-e_{kj})\,:\,1\le i\le r\}$ forms a basis if ${\mathcal {B}}$ is skew symmetric. So subspace of symmetric (resp. skew symmetric) forms is of dimension $r=\min \{r,s\}$. $\square$

Lemma 4.4

Let V be a finite dimensional vector space over ${\mathbb {F}}$, $T:V \rightarrow V$ an invertible linear operator. For $r+s=n$, $\alpha =\pm 1$, if $(x-\alpha )^r$ and $(x-\alpha )^s$, occur as elementary divisors of $\chi _T(x)=(x-\alpha )^r(x-\alpha )^s=m_T(x)$, then dimension of the space of   T-invariant bilinear forms is $n+2\cdot \min \{r,s\}$.

Proof

Without loss of generality, we can assume that $V=V_{(x-\alpha )^r}\oplus V_{(x-\alpha )^s}$. Then $\dim V=r+s=n$.

$\square$

COROLLARY 4.5

With the hypothesis as in the lemma,  if n is odd,  then the subspace of T-invariant symmetric forms is of dimension $\frac{n+1}{2}+\min \{r,s\}$ and the subspace of T-invariant skew symmetric forms is of dimension $\frac{n-1}{2}+\min \{r,s\}$.

Proof

As n is odd we have two cases:

Case 1. r odd and s even. By Corollaries 3.2 and 3.3, we have $\dim \mathcal B_{\mathrm{sym}}|_{V_{(x-1)^r}}=\frac{r+1}{2}$, $\dim {\mathcal {B}}_\mathrm{skew}|_{V_{(x-1)^r}}=\frac{r-1}{2}$, $\dim {\mathcal {B}}_\mathrm{sym}|_{V_{(x-1)^s}}=\frac{s}{2}$ and $\dim {\mathcal {B}}_\mathrm{skew}|_{V_{(x-1)^s}}=\frac{s}{2}$. So

$$\begin{aligned} \dim {\mathcal {B}}_{\mathrm{sym}}= & {} \dim B_{\mathrm{sym}}|_{V_{(x-1)^r}}+\dim B_{\mathrm{sym}}|_{V_{(x-1)^s}}+r\\= & {} \frac{r+1}{2}+\frac{s}{2}+r=\frac{n+1}{2}+\min \{r,s\} \end{aligned}$$

and

$$\begin{aligned} \dim {\mathcal {B}}_{\mathrm{skew}}= & {} \dim B_{\mathrm{skew}}|_{V_{(x-1)^r}}+\dim B_{\mathrm{skew}}|_{V_{(x-1)^s}}+r\\= & {} \frac{r-1}{2}+\frac{s}{2}+r=\frac{n-1}{2}+\min \{r,s\}. \end{aligned}$$

Case 2. r even and s odd. Again by Corollaries 3.2 and 3.3, we have $\dim \mathcal B_{\mathrm{sym}}|_{V_{(x-1)^r}}=\frac{r}{2}$, $\dim {\mathcal {B}}_\mathrm{skew}|_{V_{(x-1)^r}}=\frac{r}{2}$, $\dim {\mathcal {B}}_\mathrm{sym}|_{V_{(x-1)^s}}=\frac{s+1}{2}$ and $\dim {\mathcal {B}}_\mathrm{skew}|_{V_{(x-1)^s}}=\frac{s-1}{2}$. So

$$\begin{aligned} \dim {\mathcal {B}}_{\mathrm{sym}}= & {} \dim B_{\mathrm{sym}}|_{V_{(x-1)^r}}+\dim B_{\mathrm{sym}}|_{V_{(x-1)^s}}+r\\= & {} \frac{r}{2}+\frac{s+1}{2}+r=\frac{n+1}{2}+\min \{r,s\} \end{aligned}$$

and

$$\begin{aligned} \dim {\mathcal {B}}_{\mathrm{skew}}= & {} \dim B_{\mathrm{skew}}|_{V_{(x-1)^r}}+\dim B_{\mathrm{skew}}|_{V_{(x-1)^s}}+r\\= & {} \frac{r}{2}+\frac{s-1}{2}+r=\frac{n-1}{2}+\min \{r,s\}. \end{aligned}$$

Thus in both the cases, $\dim {\mathcal {B}}_\mathrm{sym}=\frac{n+1}{2}+\min \{r,s\}$ and $\dim {\mathcal {B}}_\mathrm{skew}=\frac{n-1}{2}+\min \{r,s\}$. $\square$

COROLLARY 4.6

With the hypothesis as in the lemma, let n be even. Then if r is odd,  the subspaces of T-invariant symmetric forms and T-invariant skew symmetric forms are of dimensions $\frac{n}{2}+1+\min \{r,s\}$ and $\frac{n}{2}-1+\min \{r,s\}$ respectively. If r is even,  they are of dimensions $\frac{n}{2}+\min \{r,s\}$ and $\frac{n}{2}+\min \{r,s\}$ respectively.

Proof

Similar. $\square$

Lemma 4.7

Let V be a finite dimensional vector space over ${\mathbb {F}}$, $T:V\rightarrow V$ a linear operator. If for $\pm 1\ne \alpha ,\,\beta \in {\mathbb {F}}$, $\alpha \beta =1$ the minimal and characteristic polynomials of T are $m_T(x)=(x-\alpha )(x-\beta )$ and $\chi _T(x)=(x-\alpha )^r(x-\beta )^s$ respectively, then dimension of the space ${\mathcal {B}}$ of T-invariant bilinear forms is 2rs.

Proof

Without loss of generality, assume that $r\le s$, $r+s=n$,

$$\begin{aligned} \displaystyle V=\oplus _{i=1}^r(V_{(x-\alpha )}\oplus V_{(x-\beta )})\oplus (\oplus _{j=1}^{s-r}V_{(x-\beta )}) \end{aligned}$$

and

where the first diagonal block is of order 2r and last is of order $s-r$.

Let $\{\displaystyle e_1,\ldots , e_n\}$ be the standard basis for V. Let $B\in {\mathcal {B}}$ be a T-invariant bilinear form on V given by $X=(x_{i,j})=(B(e_i,e_j))$.

Let us first determine $B|_{\oplus _{i=1}^{r}(V_{(x-\alpha )}\oplus V_{(x-\beta )})}$. For $1\le i,j\le 2r$,

$$\begin{aligned} x_{ij}= & {} B(e_i,e_j)=B(Te_i,Te_j)\\= & {} \left\{ \begin{array}{ll} \alpha ^2B(e_i,e_j)&{}\quad \text{ if } i \text{ and } j \text{ are } \text{ odd },\\ \alpha \beta B(e_i,e_j)&{}\quad \text{ if } i \text{ and } j \text{ are } \text{ of } \text{ distinct } \text{ parity },\\ \beta ^2B(e_i,e_j)&{}\quad \text{ if } i \text{ and } j \text{ are } \text{ even }, \end{array}\right. \end{aligned}$$

but as $\alpha ^2\ne 1$ and $\beta ^2\ne 1$, we have for $1\le i,j\le 2r$, $x_{ij}=0$ if i and j are of the same parity, whereas no condition if i and j are of different parities. So $S_1=\{e_{ij},\,\,1\le i,j\le 2r\,:\,2\not \mid i+j\}$ forms a basis for ${\mathcal {B}}|_{\oplus _{i=1}^{r}(V_{(x-\alpha )}\oplus V_{(x-\beta )})}$, with $\dim \mathcal B|_{\oplus _{i=1}^{r}(V_{(x-\alpha )}\oplus V_{(x-\beta )})}=\displaystyle {r\atopwithdelims ()1}{r\atopwithdelims ()1}+{r\atopwithdelims ()1}{r\atopwithdelims ()1}=2r^2$.

Now let us determine $\displaystyle B|_{\oplus _{i=1}^{s-r} V_{(x-\beta )}}$. For $u, v\in \oplus _{i=1}^{s-r} V_{(x-\beta )}$, $B(u,v)=B(Tu,Tv)=\beta ^2B(u,v)$, again as $\beta ^2\ne 1$, $B(u,v)=0$. Thus $\displaystyle {\mathcal {B}}|_{\oplus _{i=1}^{s-r} V_{(x-\beta )}}=0.$

The $2r\times (s-r)$ block is determined as follows: For $1\le i\le 2r$ and $1\le j\le s-r$,

$$\begin{aligned} x_{i,2r+j}=B(e_i,e_{2r+j})=B(Te_i,Te_{2r+j})=\left\{ \begin{array}{ll} \alpha \beta B(e_i,e_{2r+j})&{}\quad \text{ if } i \text{ is } \text{ odd },\\ \beta ^2B(e_i,e_{2r+j})&{}\quad \text{ if } i \text{ is } \text{ even },\end{array}\right. \end{aligned}$$

i.e., $1\le j\le s-r$, $x_{i,2r+j}=0$ if i is even. So $S_2=\displaystyle \{e_{i,2r+j},\,1\le i\le 2r,\,1\le j\le s-r\,:\,i \text{ is } \text{ odd }\}$ forms a basis for this $2r\times (s-r)$ block and its dimension is $\displaystyle {r\atopwithdelims ()1}{s-r\atopwithdelims ()1}=r(s-r)$.

Finally the $(s-r)\times 2r$ block is determined as follows: For $1\le j\le 2r$ and $1\le i\le s-r$,

$$\begin{aligned} x_{2r+i,j}=B(e_{2r+i},e_j)=B(Te_{2r+i},Te_j)=\left\{ \begin{array}{ll} \alpha \beta B(e_{2r+i},e_j)&{}\quad \text{ if } j \text{ is } \text{ odd },\\ \beta ^2B(e_{2r+i},e_j)&{}\quad \text{ if } j \text{ is } \text{ even }, \end{array}\right. \end{aligned}$$

i.e., $1\le i\le s-r$, $x_{2r+i,j}=0$ if j is even. So $S_3=\displaystyle \{e_{2r+i,j},\,1\le i\le s-r,\,1\le j\le 2r\,:\,j \text{ is } \text{ odd }\}$ forms a basis for this $(s-r)\times 2r$ block and its dimension is $\displaystyle {r\atopwithdelims ()1}{s-r\atopwithdelims ()1}=r(s-r)$.

Thus $\dim {\mathcal {B}}=2r^2+0+r(s-r)+r(s-r)=2rs$, with the disjoint union $S_1\cup S_2\cup S_3$ as a basis. $\square$

COROLLARY 4.8

Under the hypothesis of the lemma,  dimension of the subspace ${\mathcal {B}}_{\mathrm{sym}}$ of T-invariant symmetric bilinear forms is $rs=\frac{1}{2}\dim {\mathcal {B}}$.

Proof

Follows from the lemma. $\square$

COROLLARY 4.9

Under the hypothesis of the lemma,  dimension of the subspace ${\mathcal {B}}_{\mathrm{skew}}$ of T-invariant skew symmetric bilinear forms is $rs=\frac{1}{2}\dim {\mathcal {B}}$.

Proof

Similar. $\square$

Remark 4.10

If $r=s$, then there are nondegenerate T-invariant bilinear forms (symmetric as well as skew symmetric) for T as described in the lemma.

If $r=s$ and

$$\begin{aligned} T=\left( \begin{array}{lllll} \alpha &{}&{}&{}&{}\\ &{}\beta &{}&{}&{}\\ &{}&{}\ddots &{}&{}\\ &{}&{}&{}\alpha &{}\\ &{}&{}&{}&{}\beta \end{array}\right) , \end{aligned}$$

let $\sigma \in S_{2r}$ be a permutation given by

$$\begin{aligned} \displaystyle \sigma =\left\{ \begin{array}{ll} \Pi _{\small i=1,\,\,i \text{ even }}^{r}(i,\,r+i-1)&{}\text{ if } r \text{ is } \text{ even },\\ \Pi _{\small i=1,\,\,i \text{ even }}^{r}(i,\,r+i)&{}\text{ if } r \text{ is } \text{ odd }. \end{array}\right. \end{aligned}$$

Let P be the matrix representation of the permutation $\sigma$ with respect to the standard basis ($P(e_i)=e_{\sigma (i)}$). Then P gives an isomorphism of V onto V and for $B\in {\mathcal {B}}$ as obtained in the lemma, we have , where $O,\,C,\,D\in M_r({\mathbb {F}})$, O is a zero matrix. Thus for $C,\,\,D\in Gl_r({\mathbb {F}})$, B is nondegenerate. As an example $C=D=I_r$ gives rise to a nondegenerate T-invariant bilinear form. Also we can take $D=C^t\in Gl_r({\mathbb {F}})$. Then B thus obtained will be a nondegenerate T-invariant symmetric form, similarly we can take $C\in Gl_r({\mathbb {F}})$ and $D=-C^t$, then B thus obtained will be a nondegenerate T-invariant skew symmetric bilinear form.

Further, note that if $r\ne s$, we can assume $r<s$ and P as above. Then for B a T-invariant bilinear form, we have

where $\displaystyle O, C, D\in M_r({\mathbb {F}})$, $\displaystyle E, O_1\in M_{r\times s-r}({\mathbb {F}})$, $\displaystyle F, O_2\in M_{s-r\times r}({\mathbb {F}})$, $O_3\in M_{s-r}({\mathbb {F}})$, $\displaystyle O, O_1, O_2, O_3$ are zero matrices. The first row consisting of the blocks $\displaystyle O, C, E$ is of rank atmost r, the last two rows are of rank same as the column rank of the matrix $\left( \begin{array}{l}D\\ F\end{array}\right)$ which can be atmost r. Thus in all the matrix H is of rank atmost $2r<r+s$, so B cannot be nondegenerate.

5 Main results

5.1 Invariance under self-reciprocal operators

Recall that for V a n dimensional vector space over ${\mathbb {F}}$, $T\in \text {End}(V)$ is cyclic if there exists $v\in V$ such that the set $\{\displaystyle v,Tv,\ldots ,T^{n-1}v\}$ forms a basis for V.

Theorem 5.1

Let V be a vector space of dimension 2n over ${\mathbb {F}}$, $T:V\rightarrow V$ an invertible, cyclic, self-reciprocal transformation. Then the space of T-invariant bilinear forms is of dimension 2n over ${\mathbb {F}}$.

Proof

Let $m_T(x)$ be the minimal polynomial of T. Since T is cyclic $\chi _T(x)=m_T(x)$. Let $\chi _T(x)=p(x)=\sum _{i=0}^{2n}c_ix^i\in {\mathbb {F}}[x]$. As T is self-reciprocal, we have $c_0=1, c_i=c_{2n-i}$ for all $i,\,0\le i\le n$.

As T is cyclic, there exists a vector $v\in V$ such that orbit of v with respect to T spans V, i.e., $(\displaystyle e_1=v,e_2=Tv,\ldots ,e_{2n}=T^{2n-1}v)$ forms a frame for V and $T^{2n}v=-\sum _{i=0}^{2n-1}c_iT^iv$. Fix the frame ${\underline{e}}=(e_1, \ldots , e_n)$. Let the matrix of T with respect to ${\underline{e}}$ be $[T]_e$ and the matrix of B be $X=[B]_e=(B(e_i,e_j))$. Then T-invariance amounts to $[T]_e^t[B]_e[T]_e=[B]_e$. So we have

$$\begin{aligned}C=[T]_e=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0&{}0&{}0&{}\cdots &{}0&{}-c_0\\ 1&{}0&{}0&{}\cdots &{}0&{}-c_1\\ 0&{}1&{}0&{}\cdots &{}0&{}-c_2\\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\vdots &{}\vdots \\ 0&{}0&{}0&{}\cdots &{}1&{}-c_{2n-1} \end{array}\right) .\end{aligned}$$

Let $C^t=(c_{ij}^t)$ be the transpose of the matrix C, $c_{ij}^t=c_{ji}$.

$$\begin{aligned} (C^tXC)_{ij}= & {} \sum _{k,l=1}^{2n}c_{ik}^tx_{kl}c_{lj}=\sum _{k,l=1}^{2n}c_{ki}B(e_k,e_l)c_{lj}=B\left( \sum _{k=1}^{2n}c_{ki}e_k,\sum _{l=1}^{2n}c_{lj}e_l\right) \\= & {} B(Te_i,Te_j). \end{aligned}$$

So we have

$$\begin{aligned} (B(e_i,e_j))=X=C^tXC=(B(T e_i,T e_j)). \end{aligned}$$

Thus we get $x_{ij}=x_{i+1j+1}$ for all $i,j,1\le i,j\le 2n-1$. Let $B(e_1,e_i)=\alpha _i,1\le i\le 2n$ and $B(e_j,e_1)=\beta _j,1\le j\le 2n$. Clearly $\alpha _1=\beta _1$. So

$$\begin{aligned} X=\left( \begin{array}{lllll} \alpha _1 &{} \alpha _2&{} \alpha _3&{}\cdots &{}\alpha _{2n}\\ \beta _2&{}\alpha _1&{}\alpha _2&{}\cdots &{}\alpha _{2n-1}\\ \vdots &{}\ddots &{}\ddots &{}\ddots &{}\vdots \\ \vdots &{}&{}\ddots &{}\ddots &{}\vdots \\ \beta _{2n}&{}\beta _{2n-1}&{}\beta _{2n-2}&{}\cdots &{}\alpha _1 \end{array}\right) . \end{aligned}$$

Now for $1\le i\le n$, we have

$$\begin{aligned} \alpha _{2n-i+1}= & {} B(e_1,e_{2n-i+1})=B(e_2,e_{2n-i+2})=\cdots =B(e_i,e_{2n})\nonumber \\= & {} B(T e_i,T e_{2n})=B(e_{i+1},T^{2n}v)\nonumber \\= & {} B(e_{i+1},-\sum _{j=0}^{2n-1}c_je_{j+1})=-\sum _{j=0}^{2n-1}c_jB(e_{i+1},e_{j+1})\nonumber \\= & {} -\sum _{j=i}^{2n-1}c_jB(e_{i+1},e_{j+1})-\sum _{j=0}^{i-1}c_jB(e_{i+1},e_{j+1})\nonumber \\= & {} -\sum _{j=i}^{2n-1}c_jB(e_1,e_{j-i+1})-\sum _{j=0}^{i-1}c_jB(e_{i+1},e_{j+1})\nonumber \\= & {} -\sum _{j=i}^{2n-1}c_j\alpha _{j-i+1}-\sum _{j=0}^{i-1}c_jB(e_{i+1},e_{j+1}). \end{aligned}$$

(5.1)

Similarly for $1\le k\le n$, we have

$$\begin{aligned} \beta _{2n-k+1}=-\sum _{j=k}^{2n-1}c_j\beta _{j-k+1}-\sum _{j=0}^{k-1}c_jB(e_{j+1},e_{k+1}). \end{aligned}$$

(5.2)

Now for a general bilinear form B, equations (5.1) and (5.2) can be expressed as

$$\begin{aligned} \alpha _{2n-i+1}= & {} -\sum _{j=i}^{2n-1}c_j\alpha _{j-i+1}-\sum _{j=0}^{i-1}c_j\beta _{i-j+1}, \end{aligned}$$

(5.3)

$$\begin{aligned} \beta _{2n-i+1}= & {} -\sum _{j=i}^{2n-1}c_j\beta _{j-i+1}-\sum _{j=0}^{i-1}c_j\alpha _{i-j+1} \end{aligned}$$

(5.4)

respectively for $1\le i\le n$. Thus for $i=t\le n$ and $i=s\le n$, from equations (5.3) and (5.4), we get

$$\begin{aligned} \alpha _{2n-t+1}= & {} -\beta _{t+1}-\sum _{j=1}^{t-1}c_j(\beta _{t-j+1}+\alpha _{2n-j-t+1})\nonumber \\&\quad -\sum _{j=t}^{n-1}c_j(\alpha _{j-t+1}+\alpha _{2n-j-t+1})-c_n\alpha _{n-t+1} \end{aligned}$$

(5.5)

and

$$\begin{aligned} \beta _{2n-s+1}= & {} -\alpha _{s+1}-\sum _{j=1}^{s-1}c_j(\alpha _{s-j+1}+\beta _{2n-j-s+1})\nonumber \\&\quad -\sum _{j=s}^{n-1}c_j(\beta _{j-s+1}+\beta _{2n-j-s+1})-c_n\beta _{n-s+1} \end{aligned}$$

(5.6)

respectively. Thus $\displaystyle \alpha _{n+1},\ldots , \alpha _{2n}, \beta _{n+2}, \ldots , \beta _{2n}$ are expressible in terms of $\alpha _1,\ldots , \alpha _n, \beta _2, \ldots , \beta _{n+1}$.

It follows that the dimension is 2n. $\square$

Remark 5.2

Remark that ‘space of T-invariant forms = subspace of symmetric T-invariant forms $\oplus$ subspace of skew symmetric T-invariant forms’.

COROLLARY 5.3

Under the hypothesis of the theorem, the space of T-invariant symmetric bilinear forms is of dimension n.

Proof

If B is a symmetric bilinear form, then for $1\le j\le 2n$, $\alpha _j=\beta _j$ and for $1\le i\le n$ from equation (5.3), we have

$$\begin{aligned} \alpha _{2n-i+1}= & {} -\sum _{j=i}^{2n-1}c_j\alpha _{j-i+1}-\sum _{j=0}^{i-1}c_jB(e_{j+1},e_{i+1})\\= & {} -\sum _{j=i}^{2n-1}c_j\alpha _{j-i+1}-\sum _{j=0}^{i-1}c_jB(e_1,e_{i-j+1})\\= & {} -\sum _{j=i}^{2n-1}c_j\alpha _{j-i+1}-\sum _{j=0}^{i-1}c_j\alpha _{i-j+1}. \end{aligned}$$

Thus for $i=t\le n$, we have

$$\begin{aligned} \alpha _{2n-t+1}= & {} -\alpha _{t+1}-\sum _{j=1}^{t-1}c_j(\alpha _{t-j+1}+\alpha _{2n-j-t+1})\nonumber \\&\quad -\sum _{j=t}^{n-1}c_j(\alpha _{j-t+1}+\alpha _{2n-j-t+1})-c_n\alpha _{n-t+1}. \end{aligned}$$

These are the only non identically zero equations in this case. So out of $\{\displaystyle \alpha _1\cdots \alpha _{n+1}\}$ (as appearing in the theorem), $\displaystyle \alpha _1\cdots \alpha _n$ are the free scalars. Thus the set of T-invariant symmetric bilinear forms turns out to be a vector space over ${\mathbb {F}}$ of dimension n. $\square$

COROLLARY 5.4

Under the hypothesis of the theorem,  the space of T-invariant skew symmetric bilinear forms is of dimension n.

Proof

Follows from the theorem, remark and the previous corollary. $\square$

Note that there are always nondegenerate T-invariant bilinear forms for T as in the theorem. As an example, if we take for $1\le i\le 2n$, $\alpha _i=-\beta _i$, $\alpha _j=0$ for all $1\le j\le n$ and $\alpha _{n+1}\ne 0$ (the choice of this non-zero value is assured from equations (5.5) and (5.6)), then X is

$$\begin{aligned} X=\left( \begin{array}{ll} O&{}A\\ -A^t&{}O \end{array}\right) , \end{aligned}$$

where O is the zero square matrix of order n, A is the upper triangular square matrix of order n with $\alpha _{n+1}$ as entries on the main diagonal. Clearly X is skew symmetric and as $\det X=\det ^2 A=\alpha _{n+1}^{2n}\ne 0$, it is nondegenerate. Now let $Y=(C-C^{-1})^tX$, then we have

$$\begin{aligned} C^tYC= & {} C^t(C-C^{-1})^tXC=C^t(C-C^{-1})^t(C^{-1})^tC^tXC\\= & {} (C^{-1}(C-C^{-1})C)^tX=(C-C^{-1})^tX=Y, \end{aligned}$$

i.e., Y gives a T-invariant bilinear form. Now we will see that Y is symmetric. As $Y=(C-C^{-1})^tX=C^tX-(C^{-1})^tX=C^tX-(C^{-1})^tC^tXC=C^tX-XC$, we have $Y^t=X^tC-C^tX^t=-XC+C^tX=C^tX-XC=Y$. Now since by hypothesis none of $\pm 1$ is an eigenvalue of C, $C-C^{-1}$ is invertible and as X is invertible, we get Y as a nondegenerate T-invariant symmetric bilinear form.

Remark 5.5

There exist symmetric nondegenerate and skew symmetric nondegenerate forms invariant under the linear operator as in the theorem.

5.2 Invariance under indecomposable operators

Theorem 5.6

Let V be a vector space of dimension n over ${\mathbb {F}}$. If T is an indecomposable,  self-reciprocal transformation on $V$ $,$ then the space of T-invariant bilinear forms is of dimension n over ${\mathbb {F}}$.

Proof

As T is self-reciprocal, n is even. T is invertible and as T is indecomposable it is cyclic also. So by Theorem 5.1, this theorem follows. $\square$

COROLLARY 5.7

Under the hypothesis of the theorem the subspace of T-invariant symmetric (resp. skew-symmetric) forms is of dimension $\frac{n}{2}$ and this subspace contains a nondegenerate form.

Proof

Clear from the corollaries of Theorem 5.1. $\square$

Theorem 5.8

Let V be a vector space of dimension n over ${\mathbb {F}}$. If T is an indecomposable,  invertible,  non-unipotent,  non self-reciprocal transformation on V,  then the space of T-invariant bilinear forms is 0.

Proof

Since T is indecomposable, invertible, the characteristic polynomial of T is $\chi _T(x)=p(x)^d,p(x)\in {\mathbb {F}}[x]$ irred., $d\in {\mathbb {Z}}^+$. So $V=\ker (p(x)^d)$. Now due to Proposition 2.5, we can pass our investigation to ${{\bar{{\mathbb {F}}}}}$. Then we have

$$\begin{aligned} V=\oplus _{\alpha \in {{\bar{{\mathbb {F}}}}}\,\text{ a } \text{ root } \text{ of } \,p(x)}V_\alpha ,\,\text{ where }\,V_\alpha =\ker {(x-\alpha )^{dk}}, \end{aligned}$$

with

$$\begin{aligned} k=\left\{ \begin{array}{ll} 1,&{}\hbox { if char}\ {\mathbb {F}}=0,\\ (\hbox { char}\ {\mathbb {F}})^s, s\in {\mathbb {Z}}^+,&{}\text{ otherwise }. \end{array}\right. \end{aligned}$$

Let B be a T-invariant bilinear form on V. If $u\in V_\alpha$ and $v\in V_\beta$, then $B(Tu,Tv)=B(u,v)$ amounts to $B(T_su,T_sv)=B(u,v)$, i.e., $(\alpha \beta -1)B(u,v)=0$. As p(x) is nonunipotent $\alpha \ne \pm 1\ne \beta$ also as p(x) is non self-reciprocal $\beta \ne \alpha ^{-1}$. Thus we have $B(u,v)=0$. Also by the previous lemma, ${\mathcal {B}}_{V_\alpha }=0$. Therefore the space of T-invariant forms is 0. $\square$

5.3 Invariant under decomposable operators

Theorem 5.9

Let V be a finite dimensional vector space over ${\mathbb {F}},$ $T:V\rightarrow V$ an invertible linear operator. If for $p(x)\in {\mathbb {F}}[x]$ irreducible with $p(\pm 1)\ne 0$, $p(x)^r$ and $p^*(x)^s$ occur as elementary divisors of $\chi _T(x)=p(x)^rp^*(x)^s$. Then the dimension of the space of T-invariant bilinear forms is $2\cdot \deg p(x)\cdot \min \{r,s\}$.

Proof

Since $p(x)\in {\mathbb {F}}[x]$ is irreducible, all its roots are of same multiplicity, say k. So by Lemma 2.1, all roots of $p^*(x)$ are also of the same multiplicity, i.e., of k. So $\dim V=kr+ks$. Let $R=kr$ and $S=ks$. As invariances of $\chi _T$ does not change over the extensions of the field, over ${{\bar{{\mathbb {F}}}}}$, it will remain the same and over ${{\bar{{\mathbb {F}}}}}$, we have

$$\begin{aligned} \displaystyle V=V_{p(x)^rp^*(x)^s}=\oplus _{\alpha \beta =1,\,p(\alpha )=0}(V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}). \end{aligned}$$

Since $\alpha ,\,\beta \ne \pm 1$, by Lemma 4.2, we have

$$\begin{aligned} \dim {\mathcal {B}}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}=2\cdot \min \{R,S\}. \end{aligned}$$

So

$$\begin{aligned} \dim {\mathcal {B}}= & {} \sum _{\tiny \alpha \beta =1,\,p(\alpha )=0}\dim {\mathcal {B}}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}\\= & {} \sum _{\tiny \hbox { distinct roots of}\ p(x)}2\cdot \min \{R,S\}\\= & {} \hbox {} \# \, \hbox {distinct roots of} p(x)\times 2\cdot \min \{R,S\}\\= & {} \frac{\deg p(x)}{k}\cdot 2\cdot \min \{R,S\}=2\deg p(x)\cdot \min \{r,s\}. \end{aligned}$$

$\square$

COROLLARY 5.10

Under the hypothesis of the theorem,  the dimension of the subspace of T-invariant symmetric (resp. skew symmetric) bilinear forms is $\deg p(x)\min \{r,s\}$.

Proof

Again, without loss of generality, assume that $V=V_{p(x)^rp^*(x)^s}$ and $T=T|_{V_{p(x)^rp^*(x)^s}}$. Then as in the proof of the theorem, we obtain

$$\begin{aligned} \displaystyle V=V_{p(x)^rp^*(x)^s}=\oplus _{\alpha \beta =1,\,p(\alpha )=0}(V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}). \end{aligned}$$

Since $\alpha ,\,\,\beta \ne \pm 1$, by Corollary 4.3, we have

$$\begin{aligned} \dim {\mathcal {B}}_\mathrm{sym}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}=\min \{R,S\}=\frac{1}{2}\dim {\mathcal {B}}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}, \end{aligned}$$

and

$$\begin{aligned} \dim {\mathcal {B}}_\mathrm{skew}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}=\min \{R,S\}=\frac{1}{2}\dim {\mathcal {B}}|_{V_{(x-\alpha )^R}\oplus V_{(x-\beta )^S}}. \end{aligned}$$

So

$$\begin{aligned} \dim {\mathcal {B}}_\mathrm{sym}=\frac{1}{2}\dim {\mathcal {B}}=\dim {\mathcal {B}}_{skew}, \end{aligned}$$

hence by the theorem

$$\begin{aligned} \dim {\mathcal {B}}_{\mathrm{sym}}=\dim {\mathcal {B}}_\mathrm{skew}=\deg p(x)\cdot \min \{r,s\}. \end{aligned}$$

$\square$

Theorem 5.11

Let V be a finite dimensional vector space over ${\mathbb {F}},$ $T:V\rightarrow V$ a linear operator. Let the minimal and characteristic polynomials of T are $m_T(x)=f(x)f^*(x)$ and $\chi _T(x)=f(x)^rf^*(x)^s,$ $f(x)\in {\mathbb {F}}[x]$ is separable with $f(\pm 1)\ne 0$ and for $u\in {{\bar{{\mathbb {F}}}}}$ a root of f(x),  $f(u^{-1})\ne 0$. Then dimension of the space of T-invariant bilinear forms is $2\cdot \deg f(x)rs$.

Proof

As f(x) is separable, by Lemma 2.1, $f^*(x)$ is also separable and so according to the hypothesis, $m_T(x)$ has distinct roots. Hence over ${{\bar{{\mathbb {F}}}}}$, we have

$$\begin{aligned} \displaystyle V=\oplus _{\small \alpha \beta =1,\alpha \,\,\text{ a } \text{ root } \text{ of }\,f(x)}(V_{(x-\alpha )^r}\oplus V_{(x-\beta )^s}) \end{aligned}$$

and for every root $\alpha$ of f(x), we have

Let $B\in {\mathcal {B}}$ be a T-invariant bilinear form. For $\alpha ,\,\,\alpha '$ two distinct roots of f(x), if $u\in V_{(x-\alpha )^r}\oplus V_{(x-\beta )^s}$ and $v\in V_{(x-\alpha ')^r}\oplus V_{(x-\beta ')^s}$, then $u=u_1+u_2$ and $v=v_1+v_2$ with $u_1\in V_{(x-\alpha )^r}$, $u_2\in V_{(x-\beta )^s}$, $v_1\in V_{(x-\alpha ')^r}$, $v_2\in V_{(x-\beta ')^s}$ uniquely. $B(u_1,v_1)=B(Tu_1,Tv_1)$ implies $(\alpha \alpha '-1)B(u_1,v_1)=0$, as $\alpha \alpha '\ne 1$, $B(u_1,v_1)=0$. By similar arguments, we see that $B(u_i,v_j)=0=B(v_j,u_i)$ for $1\le i,j\le 2$, so $B(u,v)=0=B(v,u)$.

Thus $\dim {\mathcal {B}}=\sum _{\small \alpha \beta =1,\alpha \,\,\text{ a } \text{ root } \text{ of }\,f(x)}\dim {\mathcal {B}}|_{V_{(x-\alpha )^r}\oplus V_{(x-\beta )^s}}$. But by Lemma 4.8, $\dim {\mathcal {B}}|_{V_{(x-\alpha )^r}\oplus V_{(x-\beta )^s}}=2rs$ and we get

$$\begin{aligned} \dim {\mathcal {B}}=\sum _{\small \alpha \,\,\text{ a } \text{ root } \text{ of }\,f(x)}2rs=2rs\deg f(x). \end{aligned}$$

$\square$

COROLLARY 5.12

Under the hypothesis of the theorem,  dimension of the subspace of T-invariant symmetric bilinear forms is $\deg f(x)rs$.

Proof

Clear. $\square$

COROLLARY 5.13

Under the hypothesis of the theorem,  dimension of the subspace of T-invariant skew symmetric bilinear forms is $\deg f(x)rs$.

Proof

Similar. $\square$

Remark 5.14

If $r=s$ and T as described in the theorem, then there are nondegenerate T-invariant bilinear forms (symmetric as well as skew symmetric).

Let $B\in {\mathcal {B}}$ be a T-invariant bilinear form. Let $\deg f(x)=k$ and $\displaystyle \alpha _1, \ldots , \alpha _k$ be the distinct roots of f(x). Then for every $\alpha _i$ assuming $V_i=V_{(x-\alpha _i)^r}\oplus V_{(x-\beta _i)^r}$, we have $B_i=B|_{V_i}$ is a $T_i=T|_{V_i}$-invariant form. By Remark 4.10, for every $i,\,1\le i\le k$, we have an isomorphism $P_i:V_i\rightarrow V_i$ such that , with $O,\,C_i,\,D_i\in M_r({\mathbb {F}})$, O is a zero matrix. Now let

$$\begin{aligned} P=\left( \begin{array}{lll} P_1&{}&{}\\ &{}\ddots &{}\\ &{}&{}P_k \end{array}\right) ,\,\,\text{ then }\,\, P^tBP=\left( \begin{array}{lll} ({P_1}^tB_1P_1)&{}&{}\\ &{}\ddots &{}\\ &{}&{}({P_k}^tB_kP_k) \end{array}\right) , \end{aligned}$$

i.e.,

Thus for all $i,\,1\le i\le k$, we can choose $C_i,\,D_i\in Gl_r({\mathbb {F}})$ and get B a nondegenerate T-invariant bilinear form. Further, we can choose $C_i\in Gl_r({\mathbb {F}})$ and $D_i={C_i}^t$ (resp. $-{C_i}^t$) and get B a nondegenerate T-invariant symmetric (resp. skew symmetric) bilinear form. As an explicit example, we can choose $C_i=I_r$ and $D_i=I_r$ (resp. $-I_r$) and obtain a T-invariant nondegenerate symmetric (resp. skew symmetric) bilinear form.

Note that if $r\ne s$, by Remark 4.10, $P_i^tB_iP_i$ are degenerate for all $i,\,1\le i\le k$; hence B is always degenerate.

5.4 Infinitesimal version

DEFINITION 5.15

A linear transformation is said to be $\alpha$-self dual if its characteristic polynomial is so.

In the next theorem, we characterize the bilinear forms infinitesimally invariant under a cyclic, $\alpha$-self dual transformation $T\in \text {End}(V)$. Since T is cyclic $\chi _T(x)=m_T(x)$.

Theorem 5.16

Let V be a vector space of dimension 2n over ${\mathbb {F}},$ $T\in \mathrm{End}(V)$ cyclic,  $\alpha$-self dual,  then the space of T-invariant infinitesimal bilinear forms is of dimension 2n.

COROLLARY 5.17

With the hypothesis of the theorem,  the subspace of T-invariant infinitesimally symmetric bilinear forms is of dimension n.

COROLLARY 5.18

With the hypothesis of the theorem,  the subspace of T-invariant infinitesimally skew symmetric bilinear forms is of dimension n.

COROLLARY 5.19

Under the hypothesis of the theorem,  there exists a nondegenerate T-invariant infinitesimally symmetric (resp. skew symmetric) bilinear form.

Now let us discuss bilinear forms infinitesimally invariant under indecomposables. We consider two cases: (1) T nilpotent and (2) T nonnilpotent.

Case 1. T nilpotent. Let $S:V\rightarrow V$ be a linear nilpotent cyclic transformation. The Jordan form may be chosen as

$$\begin{aligned} D=\left( \begin{array}{llllll} 0&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad \cdots &{}\quad 0&{}\quad 0\\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad 1&{}\quad 0 \end{array}\right) . \end{aligned}$$

Then we have $\displaystyle Se_n=0,\, Se_i=e_{i+1}\, \forall \, i,\, 1\le i\le n-1$.

Theorem 5.20

Let V be a vector space of dimension n over ${\mathbb {F}}$. If $S\in \mathrm{End}\,(V)$ is an indecomposable nilpotent transformation,  then the space of infinitesimally T-invariant bilinear forms is of dimension n. If n is even (resp. odd),  this space has a nondegenerate skew symmetric (resp. symmetric) form but any symmetric (resp. skew symmetric) form.

COROLLARY 5.21

With the hypothesis as in the theorem,  if n is odd,  then the subspace of infinitesimally T-invariant symmetric forms is of dimension $\frac{n+1}{2}$ and the subspace of infinitesimal T-invariant skew symmetric forms is of dimension $\frac{n-1}{2}$.

COROLLARY 5.22

With the hypothesis as in the theorem,  if n is even,  then the subspace of infinitesimally T-invariant symmetric forms is of dimension $\frac{n}{2}$ and the subspace of infinitesimally T-invariant skew symmetric forms is of dimension $\frac{n}{2}$.

Case 2. T nonnilpotent. Again, in the light of Proposition 2.5, without loss of generality, we may assume that the base field is algebraically closed. By Theorem 15.3 (J E Humphreys, Linear Algebraic Groups, p. 98) (see also, [4]) for $T\in {\mathcal {O}}(V,B)$, with Jordan decomposition $T=T_s+N$, we have $T_s\in {\mathcal {O}}(V,B)$.

Theorem 5.23

Let V be a vector space of dimension n over ${\mathbb {F}}$. If T is an indecomposable,  $\alpha$-self dual transformation on V,  then the space of T-infinitesimally invariant bilinear forms is of dimension n over ${\mathbb {F}}$.

COROLLARY 5.24

Under the hypothesis of the theorem the subspace of T-infinitesimally invariant symmetric (resp. skew symmetric) forms is of dimension $\frac{n}{2}$. Also it contains a nondegenerate form.

Theorem 5.25

Let V be a vector space of dimension n over ${\mathbb {F}}$. If T is an indecomposable,  non-nilpotent,  non $\alpha$-self dual transformation on V,  then the space of T-infinitesimally invariant bilinear forms is 0.

Theorem 5.26

Let V be a finite dimensional vector space over ${\mathbb {F}},$ . If for $p(x)\in {\mathbb {F}}[x]$ irreducible with $p(0)\ne 0$, $p(x)^r$ and $p^-(x)^s$ occur as elementary divisors of $\chi _T(x),$ then the dimension of the space of $T|_{V_{p(x)^rp^-(x)^s}}$-invariant infinitesimally bilinear forms is $2\cdot \deg p(x)\cdot \min \{r,s\}$.

COROLLARY 5.27

Under the hypothesis of the theorem,  the dimension of the subspace of $T|_{V_{p(x)^rp^-(x)^s}}$-invariant symmetric (resp. skew symmetric) infinitesimally bilinear forms is $\deg p(x)\cdot \min \{r,s\}$.

Theorem 5.28

Let V be a finite dimensional vector space over ${\mathbb {F}},$ . Let the minimal and characteristic polynomials of T be $m_T(x)=f(x)f^-(x)$ and $\chi _T(x)=f(x)^rf^-(x)^s,$ $f(x)\in {\mathbb {F}}[x]$ is separable with $f(0)\ne 0$ and for $u\in {{\bar{{\mathbb {F}}}}}$ a root of f(x),  $f(-u)\ne 0$. Then dimension of the space of T-invariant infinitesimally bilinear forms is $2\cdot \deg f(x)rs$.

COROLLARY 5.29

Under the hypothesis of the theorem dimension of the subspace of T-invariant symmetric (resp. skew symmetric) infinitesimally bilinear forms is $\deg f(x)rs$.

Remark 5.30

If $r=s$ and T as described in the theorem, then there are nondegenerate T-invariant infinitesimally bilinear forms (symmetric as well as skew symmetric) and if $r\ne s$, there are no nondegenerate T-invariant infinitesimally forms. Explanations are similar to the invariance version.

Future scope. As a future work, the investigation of invariant forms under a group of linear transformations over a field may be taken.