The total derivative of a quadratic form

Question

Question: A quadratic form is a function $q: \mathbb{R}^{n} \rightarrow \mathbb{R}$ given by $$q(x) = \langle x, Bx\rangle$$ with $B$ a symmetrical $n \times n$ matrix. Using the definition of the total derivative, determine the total derivative of $q$ in a point $a \in \mathbb{R}^{n}$.

My attempt: We need to find a linear map with derivative $A$ s.t the following holds: $$ \lim_{h\rightarrow 0} \frac{q(a+h)-q(a)-A(h)}{||h||}=0 $$

We have \begin{align} q(a+h)-q(a) & = \langle a+h,Ba+Bh \rangle - \langle a, Ba \rangle \\ & = \langle a, Ba \rangle + \langle a,Bh\rangle + \langle h,Ba \rangle +\langle h,Bh \rangle - \langle a,Ba \rangle \\ &= \langle a,Bh \rangle +\langle h,Ba \rangle + \langle h,Bh \rangle \end{align}

So I need to find a $A$ such that the following holds. $$\lim_{h\rightarrow 0} \left( \frac{\langle a, Bh \rangle }{||h||} + \frac{\langle h,Ba \rangle }{||h||} + \frac{\langle h, Bh \rangle}{||h||}- \frac{Ah}{||h||} \right) = 0 $$

This is where I got stuck. I do not know how to proceed from here.

Please use \langle \rangle instead of < >, and replace a+Bh with a,Bh in your last equation. — Anne Bauval, May 10 '23 at 10:06
As for the answer: note that in your last equation, the third term can be canceled (its limit is $0$) and the first and second one are equal, since $B$ is symmetric. So, you can take $A=2a^TB.$ But this is certainly a FAQ. Let us look for some duplicate(s). — Anne Bauval, May 10 '23 at 10:08
See here and do similar computation.I believe it was done before. — Sourav Ghosh, May 10 '23 at 10:10
Does this answer your question? Prove that quadratic form is differentiable from the definition — Anne Bauval, May 10 '23 at 10:13
Hint: Show $\frac{\langle h,Bh\rangle}{|h|}\to0$ as $h\to0.$ — Thomas Andrews, May 10 '23 at 10:15
Also, since $B$ is symmetric, $\langle h,Ba\rangle=\langle Bh,a\rangle=\langle a,Bh\rangle.$ — Thomas Andrews, May 10 '23 at 10:19

score 0 · Answer 1 · answered May 10 '23 at 10:34

So thanks to the comments I came up with the following:

First we have that $$ \lim_{h \rightarrow 0} \frac{\langle h,Bh \rangle}{||h||} = 0 $$ So we need to find a A such that the following holds: $$ \lim_{h \rightarrow 0} \frac{\langle a,Bh \rangle + \langle h, Ba \rangle - A(h)}{||h||} = 0 $$

We can write $\langle a, Bh \rangle = a^{T}Bh$ and $\langle h, Ba \rangle = h^{T}Ba$. But $B$ is symmetrical so $h^{T}Ba = a^{T}Bh$. So $A = 2a^{T}B$. And then the above limit will be $0$.

The total derivative of a quadratic form

1 Answers1