# Let f,g be bilinear forms on a finite dimensional vector space. Show that there exist unique linear operators

Let f,g be bilinear forms on a finite dimensional vector space.

(a) Suppose g is non-degenerate. Show tha there exist unique linear operators T1, T2 on V such that f(a,b)=g(T1a,b)=g(a,T2b) for all a,b.

(b) Show that this result may not be true if g is degenerate.

My attempt:

Although I am clueless about this question, I was wondering if it has something to do with the adjoint of a linear operator. If we have a linear operator T on a hermitian space V then = and = for all v,w in V. But then how would you relate it to the other bilinear form and how would you use the fact that one of the bilinear forms is non-degenerate?

Denote the $k$-vector space in question by $V$. As $V$ is finite dimensional, $V$ and $V^* := L(V,k)$ (the dual space) have the same dimension. Define a map $\Theta_g \colon V \to V^*$ by $\Theta_g a = g(a, \cdot)$. As $g$ is non-degenerated, $\Theta_g$ is one-to-one: If $a \in \ker \Theta_g$, we have $g(a,b) = \Theta_ga(b) = 0$ for all $b \in V$, hence $a = 0$. As $\dim V = \dim V^*$, $\Theta_g$ is an isomorphism. Now define $T_1 \colon V \to V$ by $$T_1 a := \Theta_g^{-1}\bigl(f(a, \cdot)\bigr)$$ Then, for all $a, b \in V$, we have \begin{align*} g(T_1 a, b) &= \Theta_g(T_1 a)b\\ &= \Theta_g \Theta_g^{-1}\bigl(f(a,\cdot)\bigr)b\\ &= f(a,\cdot)(b)\\ &= f(a,b) \end{align*} The existence of $T_2$ is proved along the same lines.

Addendum: $T_1$ is unique, as if $T$ is any operator with $f(a,b) = g(Ta, b)$ for every $a, b \in V$, we have $$f(a,\cdot) = g(Ta, \cdot) = \Theta_g(Ta) \iff Ta = \Theta_g^{-1}\bigl(f(a,\cdot)\bigr) = T_1 a$$ Uniqueness of $T_2$ follows along the same lines.

If $g$ is denenerate, for example $g = 0$. Then $g(T_1 a, b) = 0$ for any $T_1 \colon V \to V$, that is, if $f \ne 0$, such a $T_1$ will not exist.

• But why are these two operators unique? Also, what happens when g is degenerate? – Sahiba Arora Oct 8 '15 at 17:41
• Added something. – martini Oct 8 '15 at 20:42

Hint: If $g$ is non-degenerate, then $g$ induces two isomorphisms $V \to V^*$:

• $\phi:V \to V^*$, given by $\phi(v)(y)=g(v,y)$

• $\psi:V \to V^*$, given by $\psi(v)(x)=g(x,v)$