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How to prove differentiability of $g(x)=x^TAx$?

What I've started with is the definition of differentiability:

Let $G \subset \mathbb{R}^n$ be open. $g:G \rightarrow \mathbb{R}^n$ is differentiable at $x \in G$ if exists a linear transformation $L: \mathbb{R}^n \rightarrow \mathbb{R}^m$ s.t.

$$\lim_{h \rightarrow 0} \frac{g(x+h)-g(x)-Lh}{||h||}=0$$

Now since $x^TAx$ is not a linear form, then I don't understand how can do anything with the above definition.

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    $\begingroup$ Related: math.stackexchange.com/questions/312077/… $\endgroup$ – Arthur Nov 17 '16 at 9:46
  • $\begingroup$ rewrite $x^T A x$ as $\sum_{i, j} a_{ij} x_i x_j$ $\endgroup$ – Denis Korzhenkov Nov 17 '16 at 9:47
  • $\begingroup$ @Arthur Yes, but I'm confused about what to do with $$\lim_{h \rightarrow 0} \frac{2x^TA \cdot h + h^TAh - Lh}{||h||}$$ that results from that first answer. Is $\frac{h}{||h||}$ perhaps something special? $\endgroup$ – mavavilj Nov 17 '16 at 9:52
  • $\begingroup$ Or maybe I can apply Squeeze theorem? $\endgroup$ – mavavilj Nov 17 '16 at 10:01
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Here, $g(x+h) = (x+h)^T A (x+h) = x^TAx + x^TAh + h^TAx + h^TAh$.

So that $$\frac{g(x+h)-g(x)}{||h||} = \frac{x^TAh + h^TAx + h^TAh}{||h||}$$.

Let $L(h) = x^TAh + h^TAx$. You can check that this is linear (in $h$).

Then, $$\frac{g(x+h)-g(x)-L(h)}{||h||} = \frac{h^TAh}{||h||}$$

Finally, note that $\lim_{h \to 0}\frac{h^TAh}{||h||} = 0$. Hence, the linear map $L(h) = x^TAh + h^TAx$ is the derivative of the map.

The way of picking $L$ was simple: I just took out all the linear terms of $h$ from the numerator using $L$. What remained were higher order terms, and these dominate $||h||$ as the quotient goes to zero. Hence, $L$ is very logically picked.

There are many ways of doing this problem. One is via what Denis suggested above. Another is via chain rule. But this is the elementary method.

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