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Drawing a parallel to linear Algebra.

Premise: The maximum of a determinate is when all the columns are orthogonal, given that the vectors are normalized.

Conclusion: The orthogonal trajectory should have a similar effect maximizing the wronskian. However, this does not seem to be true.

Given dy/dx = x the orthogonal trajectory would seem to be -x. However, this would show us that they are linearly dependent, and not maximized at all.

What is the definition of an "orthogonal function" as to orthogonal vectors? (functions whose tangent are orthogonal for every x seemed like a reasonable definition ). Shouldn't such functions maximize the wronskian? How would one maximize the wronskian?

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    Note that orthogonal trajectories are only defined with respect to a whole *family* of curves covering $\mathbb R^2$. As such, it doesn't make a lot of sense to speak of the orthogonal trajectory of a single function from $\mathbb R$ to $\mathbb R$. Also, the Wronskian is a *function* on $\mathbb R$ too; one can maximize a value, but how would one maximize a function?2011-01-21

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Vectors can only be orthogonal with respect to some inner-product. In that case, u is orthogonal to v when $\left \langle u,v \right \rangle=0$, where $\left \langle , \right \rangle$ denotes the inner-product. Typically, "orthogonal functions" refers to functions which are orthogonal with respect to the $L^2$ inner-product, i.e,$\left \langle f,g \right \rangle=\int_Ifgd\mu=0$.

It is evident, that two vectors which are orthogonal with respect to one inner-product need not be orthogonal with respect to another inner-product.