So I am currently studying Calculus and Linear Algebra and I came across the same concepts that is being applied in a lot of the proofs that I read for Calculus and Linear algebra but not capable of fully understanding it.
Claim: There exists a $\lambda \in \mathbb{R}$ such that $\nabla F(a)=\lambda \nabla G(a)$
So the claim above I am mentioning is one of the theorem in calculus known as the Lagrange Multiplier theorem. And it is the tool known to be solve the "constrained optimization" problem where $F$ is the function we try to maximize/minimize subject to the constraint $G=0$.
In the proofs that I read, it finished it off like this:
"$\nabla F(a) \cdot u=\lambda (\nabla G(a) \cdot u)$ where $u$ is an arbitrary unit vector.
Question: My question is how does finish off the proof when we still the $u$ vector sticking out in the proof? The text even mentions that it's because $u$ is an arbitrary unit vector so it does not matter or something along the lines of that but cannot comprehend it. Any clarification will be appreciated, and thanks in advance.
Once again, a lot of the proofs are finished off like that but it doesn't feel complete to me because we still have that $u$ sticking out. If there is a simpler example that I can understand such concept with will also be appreciated.
If
$(\nabla F(a)) \cdot u= (\lambda \nabla G(a)) \cdot u$
for all unit vectors $u$, then in particular, we can consider the unit vector $u=e^{(i)}$ which has a one in the $i$th position and 0's elsewhere. The dot product of any vector with $e^{(i)}$ gives the $i$th component of that vector.
Thus
$(\nabla F(a)) \cdot e^{(i)}=(\lambda \nabla G(a)) \cdot e^{(i)}$
for $i=1, 2, \ldots, n$.
$(\nabla F(a))_{i}=(\lambda \nabla G(a))_{i}$
for $i=1, 2, \ldots, n$.
Finally,
$(\nabla F(a))=(\lambda \nabla G(a))$.