# Find the distance between tangent space and the origin

Let $$\alpha \in\Bbb R$$ and let $$F:\Bbb R^k \to \Bbb R$$ be positive homogenous function of degree $$\alpha$$ i.e $$\forall x \neq 0$$ in $$\Bbb R^k$$ and for all $$\lambda >0$$ we have: $$F(\lambda x)=\lambda^{\alpha}F(x)$$ Assume that $$F$$ is continuously differentiable in $$\Bbb R^{k}\setminus\{0\}$$ and $$F(x)>0$$ $$\forall x \in \Bbb R^{k}\setminus\{0\}\$$ Let $$S=\{x| F(x)=1\}\$$

Now let $$a \in S$$ find the distance between the origin and the tangent space to the set $$S$$ at the point $$a$$.

What I did so far:

the tangent space to the set $$S$$ at the point $$a$$ is: $$\{x\in\Bbb R^k|\mathbf \nabla F(a) \cdot \mathbf{(x-a)}=0\}$$ so define $$g(x) = \mathbf \nabla F(a) \cdot \mathbf{(x-a)}$$ and $$f(x)=|x|^2$$ so I want to minimize $$f$$ with the constraint $$g(x) =0$$ using lagrange multipliers I have $$\nabla f(x) = \lambda \nabla g(x)$$ now $$\nabla g(x) = \nabla F(a)$$ and $$\nabla f(x) = 2(x_{1},x_{2},...,x_{k})$$ so we have $$2(x_{1},x_{2},...,x_{k})= \lambda \nabla F(a)\$$ we can use euler relation on $$F$$ and we have: $$\ \mathbf{a} \cdot \mathbf{\nabla F(a)}=\mathbf{\alpha} \cdot \mathbf{F(a)}$$. The point $$a$$ is in the set $$S$$ so we have $$F(a)=1$$. so euler relation become $$\ \mathbf{a} \cdot \mathbf{\nabla F(a)}=\alpha$$ puting all together we have: $$2(x_{1},x_{2},...,x_{k})= \lambda\alpha$$

I don't know how to continue from here. does my way untill now looks ok? if so can you give me a hint how I should continue?

• Is there a reason we use calculus here instead of linear algebra? Mar 23 at 19:23
• I saw this question when I was studying multivariable calculus Mar 23 at 20:18

Let's assume $$F(0)=0$$. Let's also assume $$\alpha\neq 0$$, otherwise we can get arbitrarily close to the origin in $$S$$ by considering $$F(ta)$$ for arbitrarily small, positive $$t$$.
Fix $$0\neq a\in S$$. As you said, $$\underset{h\to 0}{\lim} \frac{F((1+h)a)-F(a)}{h}=a\cdot \nabla F(a).$$ But also, $$\underset{h\to 0}{\lim}\frac{F((1+h)a)-F(a)}{h}=F(a)\underset{h\to 0}{\lim} \frac{(1+h)^\alpha-1}{h}=\alpha F(a)=\alpha.$$ So $$a\cdot \nabla F(a)=\alpha\neq 0.$$ Let $$N=\nabla F(a)/\|\nabla F(a)\|$$, where $$\|\nabla F(a)\|$$ denotes the length of $$\nabla F(a)$$. Since $$a\cdot \nabla F(a)\neq 0$$, $$\|\nabla F(a)\|\neq 0$$, and we aren't dividing by zero.
Then $$x\in S$$ iff $$x\cdot \nabla F(a)=\alpha$$.
Any $$x$$ can be written as $$x=x_\parallel+x\perp,$$ where $$x_\parallel=(x\cdot N)N$$ and $$x_\perp =x-x_\parallel$$. Then $$x_\parallel\perp x_\perp$$, so $$\|x\|^2=\|x_\parallel\|^2+\|x_\perp\|^2.$$ As noted above, $$x\in S$$ iff $$x\cdot \nabla F(a)=\alpha$$ iff $$x\cdot N=\alpha/\|\nabla F(a)\|$$ iff $$x_\parallel=\frac{\alpha}{\|\nabla F(a)\|}N.$$ Minimizing $$\|x\|^2=\|x_\parallel\|^2+\|x_\perp\|^2$$ as $$x_\perp$$ ranges over $$\{y:y\cdot N=0\}$$ yields the choice $$x_\perp=0$$, and the minimizer is $$x=x_\parallel = \frac{\alpha}{\|\nabla F(a)\|} = \frac{\alpha}{\|\nabla F(a)\|^2}\nabla F(a).$$
Another way to think about this: $$\nabla F(a)$$ is a normal to the tangent space. Since $$a$$ is in the tangent space, the plane is $$x\cdot \nabla F(a)=a\cdot \nabla F(a)=\alpha$$. The vector in this plane closest to the origin is obtained by "traveling along the normal" $$\nabla F(a)$$ until we hit the plane. I put "traveling along the normal" because if $$\alpha$$ is negative, we will actually want to move in the opposite direction of $$\nabla F(a)$$ to hit the plane. In any case, if a plane has normal vector $$F\neq 0$$ and is defined by $$x\cdot F=\alpha$$, the minimum distance vector to the origin has the form $$\lambda F$$ for some $$\lambda$$ (because that's traveling along the normal) and must satisfy $$\lambda F\cdot F=\alpha,$$ or $$\lambda = \alpha/\|F\|^2.$$
• You wrote: Any $x$ can be written as $$x=x_\parallel+x\perp,$$ where $x_\parallel=(x\cdot N)N$ and $x_\perp =x-x_\parallel$. I remember something like that when I was studying linear algebra,but that was a long time ago. Could you please direct me to notes or books that I could use to expand my knowledge on the subject? thank you Mar 24 at 11:45
• For this situation, we don't need to know more about $x_\parallel$ and $x_\perp$, because we give their definitions in the comment. But in general, this is just the orthogonal projection onto the subspace spanned by $N$. But any search for "orthogonal projection" should give a good background about existence, definitions, and properties. Here is one such intro. textbooks.math.gatech.edu/ila/projections.html Mar 25 at 9:09