# Proof in linear algebra/calculus

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.

• Your "claim["is completely incomprehensible if you don't define things... – DonAntonio Jan 16 at 17:08
• Okay, I'll add more details sorry about that. – javacoder Jan 16 at 17:10
• Added details, please check! – javacoder Jan 16 at 17:15
• Ignoring the specifics of the claim you are referring to... it appears your underlying question is "Why can we assert that for maps $A,B$ we have that $\forall x~(Ax = Bx)$ implies $A=B$?" To see this, suppose to the contrary that $A\neq B$. Try to show then that this would imply that there is at least one $x$ for which $Ax\neq Bx$. – JMoravitz Jan 16 at 17:17
• What else is there that I missed? Not everyone is capable of doing such things and be so aware of their math, not everyone's math is so strong. How about you stop being mean and ask and guide me for what is needed for your satisfaction to answer my question? – javacoder Jan 16 at 17:30

## 2 Answers

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,

$$(\nabla F(a))=(\lambda \nabla G(a))$$.

• But $u$ could be in a different direction from $e^{(i)}$ though, no? – javacoder Jan 16 at 17:20
• But here u is arbitrary so, it must satisfy any vector, in particular, $e^i$ too. – Mustang Jan 16 at 17:28
• @javacoder $u$ can be written as a linear combination of the $e^{(i)}$, so the result follows immediately by linearity of the dot product. – amd Jan 16 at 18:04
• The key here is in the use of the universal quantifier. $(\nabla F(a)) \cdot u=(\lambda \nabla G(a)) \cdot u$ for all $u$. The particular vectors $e^{(i)}$ are included in that "for all." – Brian Borchers Jan 16 at 18:11

one hint to understand is

in a finite dimensional space zero vector is the only vector orthogonal to all vectors

so from the last equation of your proof take the things in one side and use the above statement

• was it helpful? – Bijayan Ray Jan 16 at 17:14