Positive (semi)definiteness of a matrix Let $A(x)$ be a matrix whose entries $a_{i,j}(x)$ are continuous functions from $\mathbb{R}^n$ to $\mathbb{R}$. 
Then, if $A(x)$ is positive definite at $x = x^\star$, is $A(x)$ positive definite in some open ball around $x^\star$? I think the continuity of $a_{i,j}$ should somehow show that, but I am not sure how to prove it.  
On the other hand, if $A(x)$ is positive semidefinite, would an analgous claim be true?
 A: let $\phi_A(x) = \min_{\|h\|=1} \langle h, A(x)h \rangle$. $\phi_A$ is continuous since $\partial B(0,1)$ is compact.
Hence if $\phi_A(x)>0$ then there is some ball around $x$ such that $\phi_A$ is positive
in that ball.
Another proof would be to use the fact that the eigenvalues are continuous functions (in a suitable sense) of the entries of a matrix.
In the semidefinite case, take a simple $1 \times 1$ example $A(x) = \begin{bmatrix} x \end{bmatrix}$. Then $A(0)= \begin{bmatrix} 0 \end{bmatrix}$ is positive semidefinite, but clearly $A(x)$ is negative definite for all $x <0$.
A: You can use the characterization of positive definite matrices with the principal minors: if $A=[a_{ij}]_{\substack{1\le i\le n\\1\le j\le n}}$ is an $n\times n$ matrix, denote by $A_k$ the $k\times k$ matrix
$$A=[a_{ij}]_{\substack{1\le i\le k\\1\le j\le k}}$$
for $k=1,2,\dots,n$. Then $A$ is positive definite if and only if $\det A_k>0$ for $k=1,2,\dots,n$ (Sylvester's criterion).
Since the determinant is a continuous function and we have just a finite number of functions to deal with, the result follows.
