MHB Sturn-Liouville equation subtraction

[ognik]

So in my reading I have come across a few proofs that take two S-L eqtns like $ (p(x)u'(x))' + \lambda_u w(x)u(x)=0 $ , with distinct eigenvectors u & v and distinct eigenvalues $ \lambda_u , \lambda_v $. They multiply each eqtn by the other eigenvector and subtract, all good - seems a common approach. ie:

$ (p(x)u'(x))' + \lambda_u w(x)u(x)=0 $
- $ (p(x)v'(x))' + \lambda_v w(x)v(x)=0 $
$ = [p(vu'-uv')]' + (\lambda_u - \lambda_v)wuv=0 $

But the order when multiplying doesn't seem to matter? Ex. wvu = wuv? But u, v are vectors, so I shouldn't be able to swap their order?

 

[Ackbach]

They're technically eigenfunctions, and the multiplication happening there is good ol' real number multiplication. It's commutative and associative, so no issue swapping things around.

 

[ognik]

confirming further ...
so only with an operator/matrix eigenfunction product, does the order matter?
Does it matter in an inner product?

 

[Ackbach]

confirming further ...
so only with an operator/matrix eigenfunction product, does the order matter?

Definitely matters here, because sometimes the multiplication isn't defined one way, but it is the other way.

Does it matter in an inner product?

Only if it's complex-valued. If it's real-valued, then $\langle x|y\rangle=\langle y|x\rangle$. If it's complex-valued, then $\langle x|y\rangle=\overline{\langle y|x\rangle}$ (complex conjugate).

 

[ognik]

Awesome, last one:
What if the eigenfunctions were complex, then does the order matter? This is probably the heart of the original question ...

 

[Ackbach]

Well, complex multiplication is commutative:
$$(a+bi)(c+di)=ac-bd+(ad+bc)i=(c+di)(a+bi).$$
So, the multiplication in your first post is commutative (and associative, by the way). However, the order does matter in an inner product, if you have complex functions.