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My lecture notes say that: ''Given any bilinear form $\tau$ on $V$, there is a uniquely determined linear operator $T$ on $V$ such that $$\tau(v,w) = v \, .(Tw)$$ so once we've fixed a 'starting' bilinear form (the inner product), we can get any other bilinear form $\tau$ from this via a linear operator''.

Why is it necessary to fix a 'starting' point and what does this 'starting' point actually mean? Also is there a way of proving the above?

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can you explain the dot notation? usually one thinks of bilinear forms as homomorphisms $V\rightarrow V^*$ and not "operators" –  Blah Apr 2 '12 at 16:01
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What it means is: start with your favorite bilinear form for $V$. Then you can "translate" any other bilinear form you may encounter lying on the street into your favorite one through the use of a linear operator. You don't need to "start" by fixing one, but the point is that you can agree to have a "prefered" bilinear form, and then translate any other bilinear form into that one. –  Arturo Magidin Apr 2 '12 at 16:06
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Your starting bilinear form can't be arbitrary (e.g., it can't be 0). The meaning of the starting point is to fix a way of thinking about $V$ as a concrete vector space $F^n$. For a proof of what you ask about, see Section 2 of math.uconn.edu/~kconrad/blurbs/linmultialg/bilinearform.pdf –  KCd Apr 2 '12 at 17:08
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1 Answer 1

It is not necessary, as such, but it provides a way of seeing how the dot product form arises.

The relevant fact is that any linear functional $f: V \rightarrow \unicode{x211D}$ can be written in terms of the dot product, ie, $\exists \phi \in V$ such that $f(v) = v . \phi$. Furthermore, this $\phi$ is unique (this is important).

Now pick a $w \in V$, and consider the (single-variable) functional $u \mapsto \tau(u, w)$. Since it is linear, there exists an element of $V$ so that it can be written in terms of the dot product. Let me label this element $\phi_\tau(w)$. Then we have $\tau(u, w) = u . \phi_\tau(w)$. To finish the proof, we just need to show that the function $\phi_\tau: V \rightarrow V$ is linear.

Look at $\tau(u, \lambda w)$, for some scalar $\lambda$, then we have $\tau(u, \lambda w) = u . \phi_\tau(\lambda w)$, but since $\tau$ is linear in the second variable, it is also true that $\tau(u, \lambda w) = \lambda \tau(u, w) = u . (\lambda \phi_\tau(w))$. By uniqueness, we conclude that $\phi_\tau(\lambda w) = \lambda \phi_\tau(w)$.

In a similar manner, consider $\tau(u, w_1+w_2)$, with $w_1,w_2 \in V$. Along previous lines, we have $\tau(u, w_1+w_2) = u.\phi_\tau(w_1+w_2)$. Also $\tau(u, w_1+w_2)= u. (\phi_\tau(w_1) + \phi_\tau(w_2))$, so by uniqueness we have $\phi_\tau(w_1+w_2) = \phi_\tau(w_1) + \phi_\tau(w_2)$, and so $\phi_\tau$ is linear.

Since $\phi_\tau$ is linear, it can be written in the form $\phi_\tau(w) = T w$, where $T$ is a linear operator. Hence the expression $\tau(u, w) = u . (Tw)$.

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In physics & mechanics, tensors are often expressed in terms of the underlying representative vector (as in $\phi$ above) or representative linear map (as in $T$ above), rather than as scalar valued multi-linear functionals. This 'duality' took me a while to become comfortable with. –  copper.hat Apr 2 '12 at 16:59
    
What is the "dot product"? –  Blah Apr 3 '12 at 10:20
    
The inner product. –  copper.hat Apr 4 '12 at 5:28
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The formula $$\tau(v,w) = v \, \cdot(Tw)$$ which implicitly defines $T$ doesn't make sense if the scalar product $\cdot$ is not predefined somehow. –  Christian Blatter Jul 15 '12 at 11:14
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