Why positive definite matrix implies minimum?

Question

Suppose $f(x) = x^TAx$. Why $x^TAx >0$ implies that $f$ has a minimum?

Answer 1

I assume that $A$ is understood to be symmetric and that the underlying space is $\mathbb{R}^d$. Then there is an orthonormal basis $\left\{\mathbf{e}_i\right\}_{i=1}^d$ for $\mathbb{R}^d$, such that each $e_i$ is an eigenvector of $A$. Denote the corresponding eigenvalue by $\lambda_i$. We show that $f$ is strictly convex.

Let $0 < \mu < 1$ and consider $f(\mu x + (1-\mu)y)$ with $x \neq y$. Then $$f(\mu x + (1-\mu)y) = \left\langle \mu x + (1-\mu)y , A \left( \mu x + (1-\mu)y \right) \right\rangle = \left\langle \sum \limits_{i=1}^d \mu x_i \mathbf{e}_i + (1-\mu)y_i \mathbf{e}_i , A \sum \limits_{i=1}^d \mu x_i \mathbf{e}_i + (1-\mu)y_i \mathbf{e}_i \right\rangle \\= \sum \limits_{i=1}^d \lambda_i \left(\mu x_i + (1-\mu)y_i \right)^2$$ Since $x_i \neq y_i$ for some $i$ and $x^2$ is strictly convex $$\sum\limits_{i=1}^d \lambda_i \left(\mu x_i + (1-\mu)y_i \right)^2 <\sum\limits_{i=1}^d \lambda_i \left(\mu x_i^2 + (1-\mu)y_i^2 \right) = \mu f(x) + (1-\mu)f(y). $$ Thus $f$ is strictly convex.

Clearly as $\left\| x \right\| \to \infty$, then $f(x) \to \infty$. Thus $f$ is strictly convex "coercive" function and attains its infimum, or in other words has a unique minimizer in $\mathbb{R}^d$.