Let $A$ be a positive-definite real matrix in the sense that $x^T A x > 0$ for every nonzero real vector $x$. I don't require $A$ to be symmetric.
Does it follow that $\mathrm{det}(A) > 0$?
[Math] Does a positive definite matrix have positive determinant
linear algebramatricespositive definite
Best Answer
Here is en eigenvalue-less proof that if $x^T A x > 0$ for each nonzero real vector $x$, then $\det A > 0$.
Consider the function $f(t) = \det \left(t \cdot I + (1-t) \cdot A\right)$ defined on the segment $[0, 1]$. Clearly, $f(0) = \det A$ and $f(1) = 1$. Note that $f$ is continuous. If we manage to prove that $f(t) \neq 0$ for every $t \in [0, 1]$, then it will imply that $f(0)$ and $f(1)$ have the same sign (by the intermediate value theorem), and the proof will be complete.
So, it remains to show that $f(t) \neq 0$ whenever $t \in [0, 1]$. But this is easy. If $t \in [0, 1]$ and $x$ is a nonzero real vector, then $$ x^T (tI + (1-t)A) x = t \cdot x^T x + (1-t) \cdot x^T A x > 0, $$ which implies that $tI + (1-t)A$ is not singular, which means that its determinant is nonzero, hence $f(t) \neq 0$. Done.
PS: The proof is essentially topological. We have shown that there is a path from $A$ to $I$ in the space of all invertible matrices, which implies that $\det A$ and $\det I$ can be connected by a path in $\mathbb{R} \setminus 0$, which means that $\det A > 0$. One could use the same techniqe to prove other similar facts. For instance, this comes to mind: if $S^2 = \{(x, y, z) \mid x^2 + y^2 + z^2 = 1\}$ is the unit sphere, and $f: S^2 \to S^2$ is a continuous map such that $(v, f(v)) > 0$ for every $v \in S^2$, then $f$ has degree $1$.