Symmetric Positive Definite and Gradient Proof

Question

I have the function $f(x)=\frac {1}{2} \mathbf x^T Q \mathbf x - \mathbf b^T \mathbf x$ where $Q$ is symmetric.

I'm trying to show that solving $\nabla f(\mathbf x) = 0$ is equivalent to solving $Q \mathbf x = \mathbf b$

I've already proved that if $Q$ is positive definite, then $Q$ is invertible, and then $Q^{-1}$ is also positive definite. Any function that is symmetric positive definite is convex. Therefore $f$ is convex.

A convex function is guaranteed to have a local minimum which is also a global minimum which is why $\nabla f(\mathbf x) = 0$ is true.

How do I show the equivalence in an elegant way though?

Answer 1

In components, the assert is straightforward: $$ f(x) = \frac{1}{2}\sum_{ij} Q_{ij} x_ix_j - \sum_i b_i x_i. $$

So (using $Q$'s symmetry): $$ \frac{\partial f}{\partial x_k} = \frac{1}{2}\sum_{i} Q_{ik} x_i + \frac{1}{2}\sum_{j} Q_{kj} x_j - b_k = \sum_j Q_{kj} x_j - b_k , $$

which are exactly the components of the vector: $$ \nabla f = Qx - b. $$

Answer 2

Just compute $\nabla f$ directly. Because $Q$ is symmetric, you get $\nabla f(\mathbf x) = Q\mathbf x - \mathbf b$.