Question about differentiability in $\mathbb{R}^n$

Question

I found this peculiar problem:

Let $A\in \mathcal{M}_n(\mathbb{R})$ be a symmetric matrix, $B\in \mathcal{M}_{1\times n}(\mathbb{R})$ and let $C$ be a real constant. Prove that $p:\mathbb{R}^n\to \mathbb{R}$ defined as:$$p(\mathbf{x})=\langle A\mathbf{x},\mathbf{x}\rangle +B\mathbf{x}+ C$$is differentiable in $\mathbb{R}^n$.

Clearly $B\mathbf{x} +C$ is differentiable but the inner product involving $A$ is causing me some trouble. I don't get why the condition of $A$ being symmetrical is set, and I could extend the inner product in terms of $A$ to prove differentiabilty by justifying it's a $C^1 (\mathbb{R}^n)$ function; but it seems out of place. I wonder if there's an inequality (Cauchy-Schwarz will help?) I could use to prove that the inner product is differentiable by definition, or some other quicker/fancier way.

Answer 1

You are right, the condition of $A$ being symmetrical is useless here.

The inner product is bilinear and continuous (as is any bilinear map in finite dimension, but here you can indeed use Cauchy-Schwarz inequality to prove it).

Any continuous bilinear map $B$ is smooth and $$DB(a,b)(h,k)=B(a,k)+B(h,b).$$

Answer 2

It is true for each matrix $A= (a_{ij})$. The $i$-th component of $A \mathbf x$ is $\sum_{j=1}^n a_{ij}x_j$ and therefore we get. $$\langle A \mathbf x, \mathbf x \rangle = \sum_{i=1}^n(\sum_{j=1}^n a_{ij}x_j ) x_i = \sum_{i,j=1}^n a_{ij}x_jx_i .$$ Clearly all partial derivatives exist and are continuous, and therefore $\langle A \mathbf x, \mathbf x \rangle$ is differentiable.

Answer 3

$p(\mathbf{x})$ is a polynomial function of the coordinates of $\mathbf{x}$. Polynomials are differentiable. I think that is all you need to say. I agree that the condition that $\mathbf{A}$ is symmetrical is irrelevant.