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I'm reading Shilov's Linear Algebra. I had a certain confusion with bilinear forms: Suppose someone gives me a bilinear form without telling me which basis it is in, how can we know what is the basis of it? I have made two questions trying to understand this:

Q1: Can we have $2$ maps for any bilinear forms $A$ given by:

$$A(e_i,e_k)\mapsto a_{ik} \quad A(f_i,f_k)\mapsto b_{ik}$$ With two different basis $e_i,f_i$ such that $a_{ik}=b_{ik}$?

I guess the answer is no, but I may have missed something. When I tried to suppose that the answer is yes, I found that this would force us to have $e_i= f_i$ for all $i$ (It seems that there is an exception when the dimension is $1$ or $2$). So it seems (assuming I didn't made a mess) that two different basis can't have the same given maps.

Q2: Can we have $4$ maps for any bilinear forms $A$:

$$A(e_i,e_k)\mapsto a_{ik} \quad A(f_i,f_k)\mapsto b_{ik} \\ A(g_i,g_k)\mapsto c_{ik} \quad A(h_i,h_k)\mapsto d_{ik}$$

In $4$ different basis $e_i,f_i,g_i,h_i$ such that $a_{ik}=c_{ik}$ and $b_{ik}=d_{ik}$?

I have tried to show it with a very rudimentary way - basically expanding with the bilinear form properties but it doesn't seems to be very useful. I had the impression that the basis one chooses is unimportant basically from these sources:

  • Some exercises give me a matrix of coefficients $a_{ik}$ and ask me to find the matrix of coefficients $b_{ik}$ in another basis. The second basis is given but the first one is missing.

  • Most of the texts I saw jump from the description of a bilinear form to the change of matrix when one does a change of basis, the idea that a certain basis must have a fixed/limited set of coefficients seems to be missing. Although I'm not sure if there is some subtlety I missed.

  • A bilinear form is a map $A: \Bbb{V}\times \Bbb{V} \to \Bbb{K}$. Given two vectors $x=\sum_{i=1}^{n} a_ie_i$ and $y=\sum_{j=1}^{n}a_je_j$ we have:

$$A(x,y)=A\left(\sum_{i=1}^{n} a_ie_i, \sum_{j=1}^{n}a_je_j\right)=\sum_{i=1}^{n} \sum_{j=1}^{n}a_ia_j A(e_i,e_j)$$

This seems to point out that the maps from $A(e_i,e_j)$ of the basis can be made in an arbitrary way to elements in $\Bbb{K}$, when we do a change of basis, we have:

$$f_v=\sum_{i=1}^{n}b_i^ve_i$$

And the coefficients of the new matrix are going to be:

$$b_{ik}=A(f_i,f_k)=A\left(\sum_{v=1}^{n}b_v^i e_v,\sum_{w=1}^{n}b_w^k e_w \right)=\sum_{v=1}^{n}\sum_{w=1}^{n}b_v^i b_w^k A(e_v,e_w)$$

And again, we have the coefficients $A(e_v,e_w)$ without ever mentioning what was the original basis $e_i$.

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    $\begingroup$ For your first question the zero form $A\equiv 0$ provides a positive example where the coefficients are the same in any basis $\endgroup$ Jul 14, 2017 at 1:38
  • $\begingroup$ @YousufSoliman Yes, I thought about taking this case into account but I may have confused and mixed the transformation with the dimension. $\endgroup$
    – Red Banana
    Jul 14, 2017 at 1:39

1 Answer 1

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You can define a Bilinear form with no particular basis in mind. Or based on one....

If you have a matrix $C = \{c_{ij}\}$ and a basis $\{ v_1,v_2,\ldots,v_n\}$ , you can define a bilinear form $Q(v_i,v_j) = c_{ij}$ and using the bilinearity.

If you have a Bilinear form $Q$ and you pick a basis $\{v_1,v_2,\ldots,v_n\}$ you can get a matrix $C$ with $c_{ij}=Q(v_i,v_j)$. This using this $C$ above will give you $Q$ back as a bilinear form, but you don't need the basis unless you are starting from the matrix.

So I think the answer to both your questions is "yes," but you seem to be using the same letter to express different bilinear forms, so I might be confused.

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  • $\begingroup$ No, $A$ express only one bilinear form. $\endgroup$
    – Red Banana
    Jul 14, 2017 at 1:45
  • $\begingroup$ Your question says 2/4 different maps. Also, your second question is your first question asked twice. If you are in $\mathbb R^n$ these questions are quadratic equations. As a non-trivial example, the dot-product is a bilinear from in $\mathbb R^n$, and any orthonormal basis of $\mathbb R^n$ gives the same coefficient matrix... namely the identity. $\endgroup$ Jul 14, 2017 at 1:51
  • $\begingroup$ In my second question I am asking regardless of the basis and bilinear form. Mr. Soliman already showed an example of transformation related to my first question, but it seems to be only a trivial case. $\endgroup$
    – Red Banana
    Jul 14, 2017 at 1:53
  • $\begingroup$ I have corrected the question. $\endgroup$
    – Red Banana
    Jul 14, 2017 at 4:28

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