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If the relation $a_{ij}u^iu^j=0$ holds for all vectors $u^i$ such that $u^i\lambda_i=0$ where $\lambda_i$ is a given covariant vector, show that
$$a_{ij}+a_{ji}=\lambda_iv_j+\lambda_j v_i$$ where $v_j$ is some covariant vector.

I am completely stuck on this problem. How can I solve this problem?

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1 Answer 1

Let me give you some hints instead of a complete solution. Part of the problem here is to figure out what $v$ is supposed to be. In order to do this you might try guessing and looking at some examples and special cases. Thus, first suppose your coefficients are diagonal, i.e. $a_{ij} = 0$ unless $i=j$. Suppose you have somehow already found $v$ and therefore you have the relation $$a_{ii} = \lambda_iv_i.$$ Then it follows that $v_i = \lambda^ia_{ii}.$ Removing the diagonal assumption on the coefficients might make you then guess that $$v_i = \lambda^ka_{ki}.$$ Now you can just check whether or not your identity holds. In order to do this, first pick some vector $u$ so that $u^i\lambda_i = 0$. Does the identity hold for this $u$? Next, since $u$ is orthogonal to $\lambda$, by linearity it suffices (why?) to check that your identity holds for $\lambda$, i.e. you want to see whether or not the equality $$(a_{ij} + a_{ji})\lambda^i\lambda^j = (\lambda_iv_j + \lambda_jv_i)\lambda^i\lambda^j$$ is valid. If it is, then your problem is complete!

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