# How can I know what basis a bilinear form is originally in?

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 $$e_i$$.

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

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.

• No, $A$ express only one bilinear form. Jul 14, 2017 at 1:45
• 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. Jul 14, 2017 at 1:51
• 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. Jul 14, 2017 at 1:53
• I have corrected the question. Jul 14, 2017 at 4:28