Show that the set {$v: v^{T}A v \le 1$} is bounded iff $A$ is a positive definite matrix

Question

Show that the set {$v: v^{T}A v \le 1$} is bounded iff $A$ is a positive definite matrix.Where $v$ denotes a column vector.

I more or less get the if part but I have no clue how to approach the only if part. How can we infer so surely that $A$ has to be positive definite?

Answer 1

For the only if part: suppose that $w\neq 0$ is such that $w^TAw\leq 0$, then $\alpha w\in\{v:v^TAv\leq 1\}$ where $\alpha>0$ can be arbitrarily large.

Answer 2

I assume that $A$ is symmetric (or consider $(A+A^\top)/2$).

Consider the eigendecomposition of $A$: \begin{align} A = U^\top\Lambda U \end{align} where $U$ is orthogonal matrix (by spectral theorem). Then the set becomes \begin{align} \{v: v^\top A v \leq 1\} = \{v: v^\top U^\top\Lambda U v \leq 1\}=\{U^\top w: w^\top \Lambda w \leq 1\}. \end{align} Since the map $w \mapsto U^\top w$ preserves the distance, \begin{align} \{v: v^\top A v \leq 1\} \text{ is bounded} \Leftrightarrow \{ w: w^\top \Lambda w \leq 1\} \text{ is bounded}. \end{align} Since $\Lambda$ is diagonal, it is easy to verify that \begin{align} \{ w: w^\top \Lambda w \leq 1\} \text{ is bounded} \Leftrightarrow \text{diagonal entries of } \Lambda \text{ are positive}, \end{align} which concludes the proof ($\Leftarrow $ part is obvious and $\Rightarrow $ can be proved by considering contrapositive).