Show there exists $T$* such that $F(T(u), v) = F(u, T^{*}(v))$

Question

I'm having trouble with the following problem:

Let V be a vector space of finite dimension and F a non-degenerate symmetric bilinear form on V. Show that if $T \in L(V,V)$, then there exists $T$* $\in L(V,V)$ such that $F(T(u), v)= F(u, T^{*}(v))$ $\forall u,v \in V$.

I know that because F is symmetric I will be able to pass something in the first argument to the second, however, I'm having trouble thinking which could be the $T$* that works. In my course we had been using $T$* as the transpose conjugate, but I think here it isn't the case that one works. Also, I can´t figure out where to use the hypothesis F is non-degenerate.

Thanks in advance for your help.

Answer 1

Let $V^\prime$ be the dual space of $V$.

Since $F$ is non degenerate it is true that the map \begin{align*} J : V &\longrightarrow V^\prime \\ v &\longmapsto ( w \mapsto F(v,w)) \end{align*} is a linear isomorphism.

Let $T^\prime$ be the algebraic adjoint of $T$, defined by: \begin{align*} T^\prime : V^\prime &\longrightarrow V^\prime \\ v^\prime &\longmapsto \big(w \mapsto v^\prime (T(w)) \big). \end{align*} $T^\prime$ is a well defined linear map.

Now consider the following diagram $$ \require{AMScd} \begin{CD} V @. @. V @>J>> V^\prime \\ @VTVV @. @AT^* AA @AAT^\prime A \\ V @. @. V @>J>> V^\prime , \end{CD} $$ Since $J$ is an isomorphism there is a unique linear map $T^*$ that makes the above diagram commute. For the diagram to commute it is necessary (and sufficient) that $$T^* = J^{-1} T^\prime J.$$

For any $v,w \in V$ we have by definition: \begin{equation} \big(J(T^*(v))\big)(w) = F (T^*(v) ,w ) \end{equation} and \begin{equation} \big(T^\prime( J (v)) \big)(w)= F(v,T(w)). \end{equation} Now since the diagram commutes: \begin{equation} F(v,T(w)) = \big(T^\prime( J (v)) \big)(w) = (J(T^*(v))(w) = F (T^*(v) ,w ) . \end{equation} Which is the desired Identity (after using the symmetry of $F$). Notice that we have not only shown existence, but uniqueness as well.

Answer 2

Here is a brutal-force solution using coordinates.

Let $n = \dim V$, and fix a basis $\mathfrak{B} = (e_1, \ldots, e_n)$ of $V$. Then for each $v \in V$, we write the coordinate vector of $v$ by

$$ [v] = (v_1, \ldots, v_n)^{\top}, $$

where $v_i$'s are so that $v = \sum_{i=1}^{n} v_i e_i$. Also, define the $n \times n$ matrices

1. $[F] = [F_{ij}]$ by $ F_{ij} = F(e_i, e_j) $, and

2. $[T] = [T_{ij}]$ by $Te_j = \sum_{i=1}^{n} T_{ij}e_i $.

Then for any $T, S \in \mathcal{L}(V, V)$, we get

\begin{align*} F(u, v) &= F \biggl( \sum_i u_i e_i, \sum_j v_j e_j \biggr) = \sum_{i, j} u_i F_{ij} v_j = [u]^{\top} [F] [v], \\ F(Tu, v) &= F \biggl( \sum_{ik} T_{ki} u_i e_k, \sum_j v_j e_j \biggr) = \sum_{ij} \biggl( \sum_k T_{ki} F_{kj} \biggr) u_i v_j = [u]^{\top} [T]^{\top} [F] [v] , \\ F(u, Sv) &= F \biggl( \sum_{i} u_i e_i, \sum_{jk} S_{kj} v_j e_k \biggr) = \sum_{ij} \biggl( \sum_k F_{ik} S_{kj} \biggr) u_i v_j = [u]^{\top} [F] [S] [v]. \end{align*}

Since $F$ is non-degenerate, $[F]$ is invertible. Using this and defining $T^* \in \mathcal{L}(V, V)$ by

$$ [T^*] = [F]^{-1}[T]^{\top}[F], $$

then it follows that $F(Tu, v) = F(u, T^* v)$ for all $u, v \in V$.