A smoothly varying family of positive-definite matrices Consider a smoothly varying family of matrices $(g_{ij}^{t})$ where $0\leq t\leq1$. For all $0\leq t\leq1$, $\det(g_{ij}^{t})>0$ and for $t=0$, the matrix $(g_{ij}^{t})$ is positive definite. How could i conclude that $(g_{ij}^{t})$ is positive definite for any $t\in [0,1]$?
Thanks a lot for your time. 
 A: To the matrix $(g_{ij} ^t)$ one may associate the billinear form given by $g^t (u, v) = \sum \limits _{i,j} g_{ij} ^t u_i v_j$. Let $s = \inf \{s \in (0,1] \mid (g_{ij} ^s) \text{ is not positive-definite} \}$; this means that there exist a vector $u \ne 0$ such that $g^s (u, u) < 0$. Clearly $s > 0$ because $(g^0 _{ij})$ is positive-definite, so the interval $(0, s)$ is not empty.
Note that $g^0 (u, u) \ge 0$ by positive-definiteness; the fact that $\det (g^0 _{ij}) > 0$ means that $(g^0 _{ij})$ is not degenerate, therefore the inequality is strict: $g^0 (u, u) > 0$.
Consider the function $f(t) = g^t (u,u)$ for $t \in [0,s]$. Since $t \mapsto (g^t _{ij})$ is smooth, so will be $f$; in particular, it will be continuous. Note that $f(0) > 0$ and $f(s) < 0$; then, by continuity, there exist $r \in (0, s)$ with $g^r (u, u) = f(r) = 0$. But this means that $(g^r _{ij})$ is degenerate, which is a contradiction. Therefore, the set $\{s \in (0,1] \mid (g_{ij} ^s) \text{ is not positive-definite} \}$ must be empty, so $(g^t _{ij})$ is positive-definite $\forall t \in [0,1]$.
A: (Assuming $g^t$ is symmetric and real) the eigenvalues of $g^t$ are real numbers, and they are roots of a characteristic polynomial  whose coefficients vary continuously with $t$. Speaking loosely, the roots therefore also vary continuously with $t$, in  a sense that can be made precise using e.g. complex analysis methods. When $t=0$ these roots are all positive. If at some later time any root were  negative, there would be a time at which at least one root was zero. At that time $det(g^t)=0$. 
