Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a real symmetric positive definite $n \times n$ matrix. Then there exists some orthogonal matrix $O$ and diagonal matrix $D$ such that $O^T D O = A$, where $O$ has entries the eigenvectors of $A$ and $D$ has entries the real eigenvalues of $A$.

Let $x_0$ be some point in $\mathbb{R}^n$, and consider the vector $y = x_0 + O(x-x_0)$.

Then if I have any real twice-differentiable function $u: \mathbb{R}^n \to \mathbb{R}$, why can I write $$u_{x_i} = \sum_{k=1}^n u_{y_k} o_{ki}$$ and $$u_{x_i x_j} = \sum_{k,l=1}^n u_{y_k y_l} o_{ki} o_{lj}$$ where $o_{ki}$ is the entry in the $k$-th row and $i$-th column of the matrix $O$?

Also, what should be the geometric picture I have when I think about this vector $y$, and what part, if any, does the specific choice of $x_0$ play in these calculations?

share|improve this question

1 Answer 1

This has nothing to do with $O$ being orthogonal. If $O$ is any matrix at all, the first equation is true. By the chain rule, we have

$$\frac{\partial u}{\partial x_i} = \sum_k \frac{\partial u}{\partial y_k}\frac{\partial y_k}{\partial x_i}.$$

What is $\frac{\partial y_k}{\partial x_i}$? Well, since $x_0$ is constant and $O$ is constant, we have $Ox_0$ constant as well. When taking derivatives, we can ignore the constant pieces (so $x_0$ plays no role at all). We have $ y_k = (Ox)_k + (x_0)_k + (Ox_0)_k = \sum_j o_{kj}x_j + (x_0)_k +(Ox_0)_k$. Taking the partial with respect to $x_i$ gives $\frac{\partial y_k}{\partial x_i} = o_{ki}$.

Plugging this into the above formula for $\frac{\partial u}{\partial x_i}$ gives the first result.

The second result follows from the first result after noting that $o_{ki}$ is constant and replacing $u$ in the previous derivation with $u_{x_i}$.

share|improve this answer
    
Thanks for the quick response! I have one question about what you wrote. Why is the expression on the left not $\frac{\partial u \circ y}{\partial x_i}$ instead of just $\frac{\partial u}{\partial x_i}$? –  user1736 Dec 21 '11 at 3:35
    
@user1736: If you were to write out the left side, it would be $\frac{\partial u(y(x))}{\partial x_i}$, but most people write it as I did. –  Jason DeVito Dec 21 '11 at 14:42
    
Oh, I see. I just saw the notation in wikipedia too. Those two expressions aren't generally the same function though, right? –  user1736 Dec 21 '11 at 19:18
    
@user1736: Well, interpreted literally, you're right - the domain of what I wrote is "domain of y's" while the domain of the proper form is "domain of x's", and these domains need not be the same.. But $\frac{\partial u}{\partial x_i}$ only makes sense (due the domain issues) if "u", in this context, means "u$\circ$ y". In short, it may be shoddy notation technically, but in practice there's rarely any cause of confusion (at least, once you get used to it) and it's used quite commonly. –  Jason DeVito Dec 21 '11 at 19:27
    
Yea, sorry for the trouble but I guess I am still not used to it. So the only difference between the two partial derivatives is the extra mapping that changes the domains? As an example, if I look at the set $\{y: \frac{\partial u(y)}{\partial x_i} >0\}$, then is $\{x: \frac{\partial u(y(x))}{\partial x_i} > 0\}$ just the set of $x$ that $y$ maps into the first set? –  user1736 Dec 21 '11 at 20:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.