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I learned today that the set of symmetric bilinear forms (SBF) of the form $\sigma: V\times V \rightarrow F$ is in one-to-one correspondence with the set $\mathcal{S}_n(F)$ of $n$ by $n$ symmetric matrices, where $V$ is a finite-dimensional vector space over an infinite field $F$. But by a theorem I learned previously, $M_n(F)$ (of which $\mathcal{S}_n(F)$ is a subset) is in one-to-one correspondence with $\mathcal{L}(V)$, the set of all linear operators on $V$. In both cases, the correspondence is made between the linear operator/SBF and its matrix representation.

Then, presumably, with respect to a particular basis $B$ of $V$, a symmmetric matrix $A \in \mathcal{S}_n(F)$ will represent both a linear operator $L$ and a SBF $\sigma$. Besides having the same matrix representation (which maybe is as 'deep' as it gets), is there anything else I can say regarding the connection between the $L$ and $\sigma$ which satisfy $[L]_B = [\sigma]_B$?

Thanks in advance!

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Is $V$ finite dimensional? –  Michael Albanese Oct 12 '12 at 3:11
    
Yes, sir! I'll edit the original post. –  Bachmaninoff Oct 12 '12 at 3:11
    
Although, if possible, could you let me know how that would change things? We've only worked with finite-dimensional vector space, and my instructor says to just ignore the infinite-dimensional cases (which I know nothing about) for the purposes of the course. –  Bachmaninoff Oct 12 '12 at 3:14
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The difference is that for a finite-dimensional vector space $V$, there is an isomorphism between $V$ and $V^*$, but this is no longer true for infinite-dimensional vector spaces. –  Michael Albanese Oct 12 '12 at 3:23
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2 Answers 2

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I will outline how to go from a symmetric bilinear form to a linear operator.

Given a symmetric bilinear form $\sigma : V \times V \to F$, consider the map $L' : V \to V^*$ given by $L'(v) = \sigma(v, \cdot)$. It is not a linear operator as the codomain is $V^*$ not $V$. However, as $V$ is finite dimensional, $V^*$ is isomorphic to $V$; let $\phi : V^* \to V$ be an isomorphism. Now it is not hard to check that $L = \phi\circ L'$ defines a linear operator on $V$.


It is worth noting that the isomorphism between $V$ and $V^*$ is not canonical, you need to choose a basis for $V$ in order to define one.

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Thank you for the response! But what is $V^*$? I notice that it shows up in both your answer and Yury's answer. –  Bachmaninoff Oct 12 '12 at 4:46
    
It is the dual of $V$, the vector space of all linear maps $V \to F$. –  Michael Albanese Oct 12 '12 at 7:31
    
Thanks, Michael. I haven't learned this in class, but I'm going to read up on it now. In the meantime I'll leave the question open and see if there are any more answers/comments. From your conclusion (I still need to make sure I understand everything in between!), this looks like what I may have been searching for. Again, thank you! –  Bachmaninoff Oct 13 '12 at 3:25
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Every bi-linear form $\sigma:V\times V \to F$ defines a linear operator $A$ from $V$ to $V^*$; $A$ maps vector $v\in V$ to linear functional $l_v$ such that $l_v(u) = \sigma(u,v)$. The correspondence between $\sigma$ and $A$ is “canonical” — it doesn't depend on the basis.

Now suppose that we fix a basis $e_1, \dots, e_n$ in $V$. Define a basis $e_1^*, \dots, e_k^*$ in $V^*$ as follows: $e_i^*(e_i) = 1$ and $e_i^*(e_j) = 0$ if $j\neq i$. Let $\varphi$ be the map from $V^*$ to $V$ that sends $e_i^*$ to $e_i$. Then the operator $[L]_{B}$ is equal to $\varphi A$. Note that unlike the definition of $A$, the definition of $\varphi$ depends on the choice of the basis in $V$.

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Thanks, Yury! As you might see from my comments on Michael's answer, I need to teach myself some basic stuff about what $V^*$ before I can make meaning out of this! From the looks of it, though, your answers are very similar so I'll use them both to help me. Thanks! –  Bachmaninoff Oct 13 '12 at 3:27
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