Does a positive definite matrix have positive determinant? 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 $\mathrm{det}(A) > 0$?
 A: Let $A\in\mathbb{R}^{n\times n}$ such that $x^TAx\geq 0$ for any $x\in\mathbb{R}^n$. 


*

*The real eigenvalues of $A$ are non-negative.
This follows simply from the fact that to a real eigenvalue $\lambda$ of $A$, you can choose a real eigenvector $x\neq 0$. Hence, $0\leq x^TAx=\lambda x^Tx$ implies that $\lambda\geq 0$.

*The complex eigenvalues of $A$ have non-negative real parts.
Assume that $\lambda\in\mathbb{C}$, $\lambda=\xi+i\eta$, $\xi,\eta\in\mathbb{R}$ is an eigenvalue of $A$ with an associated eigenvector $x=u+iv$, $u,v\in\mathbb{R}^n$. From $Ax=\lambda x$, we obtain by equating real and imaginary part of the equality the following relations:
$$
Au=\xi u-\eta v, \quad Av=\eta u+\xi v.
$$
Premultiplying the first by $u^T$ and the second by $v^T$ gives
$$
0\leq u^TAu=\xi u^Tu-\eta u^Tv, \quad 0\leq v^TAv =\eta v^Tu+\xi v^Tv.
$$
Since $u^Tv=v^Tu$, we get by summing the two together that
$$
0\leq u^TAu+v^TAv=\xi(u^Tu+v^Tv).
$$
As before, this implies that $\xi\geq 0$ and hence $\mathrm{Re}(\lambda)\geq 0$.
To summarize, we have the following statement (including the case with a strict inequality, which follows easily from the proofs above):

Let $A\in\mathbb{R}^{n\times n}$ such that $x^TAx\geq 0$ for all $x\in\mathbb{R}^{n}$ ($x^TAx>0$ for all non-zero $x\in\mathbb{R}^n$). Then the eigenvalues of $A$ have non-negative (positive) real parts.

Now since the determinant of $A$ is the product of the eigenvalues of $A$, it is:


*

*non-negative if $x^TAx\geq 0$ for all $x\in\mathbb{R}^{n}$,

*positive if $x^TAx>0$ for all non-zero $x\in\mathbb{R}^{n}$.


SLIGHTLY LONGER NOTE: 
When determining the sign of the determinant, we do not need to care much about the complex eigenvalues and avoid thus the second item above about the real parts of the complex spectrum. Assume that $\lambda_1,\ldots,\lambda_k\in\mathbb{R}$ and $\lambda_{k+1},\bar{\lambda}_{k+1},\ldots,\lambda_{p},\bar{\lambda}_{p}\in\mathbb{C}\setminus\mathbb{R}$ are the $n$ eigenvalues of $A$ (such that $k+2(p-k)=n$). The determinant of $A$ is given by
$$\tag{1}
\det(A)=\left(\prod_{i=1}^k\lambda_i\right)\left(\prod_{i=k+1}^p\lambda_i\bar{\lambda_i}\right)=\left(\prod_{i=1}^k\lambda_i\right)\underbrace{\left(\prod_{i=k+1}^p|\lambda_i|^2\right)}_{\geq 0}
$$
The determinant is hence equal to the product of the real eigenvalues times something non-negative. 
Hence for the case $x^TAx\geq 0$ for all real $x$, one just needs to show that the real eigenvalues of $A$ are non-negative (the first item above) in order to arrive at the conclusion that $\det(A)\geq 0$. 
For the case $x^TAx>0$ for all nonzero real $x$, the analog of the first item above shows that the real eigenvalues are positive and we just need to show that the non-negative term in (1) is actually positive. This can be made simply by showing that $A$ is non-singular which implies that there is no zero eigenvalue (real or complex). Hence assume that $x^TAx>0$ for all nonzero real $x$ and $A$ is singular. Therefore, there exists $y\in\mathbb{R}^n$, $y\neq 0$, such that $Ay=0$. But $0<y^TAy=y^T0=0$ (contradiction). Consequently, $\det(A)>0$.
A: 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 $1$.
