If eigenvalues are positive, is the matrix positive definite?

I think this is false. Let $A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle < 0$.

The point is that the matrix can have all its eigenvalues strictly positive, but it does not follow that it is positive definite.

This question does a great job of illustrating the problem with thinking about these things in terms of coordinates. The thing that is positive-definite is not a matrix $M$ but the quadratic form $x \mapsto x^T M x$, which is a very different beast from the linear transformation $x \mapsto M x$. For one thing, the quadratic form does not depend on the antisymmetric part of $M$, so using an asymmetric matrix to define a quadratic form is redundant. And there is no reason that an asymmetric matrix and its symmetrization need to be at all related; in particular, they do not need to have the same eigenvalues.

As posed, the answer to the question is no, if $\mathbf A$ is not symmetric. Counterexample:

$$\mathbf A = \begin{pmatrix} 7 & 1 \\ -20 & -2\end{pmatrix}$$

with positive eigenvalues $3$ and $2$. $\mathbf A$ is not positive definite, that is, $\mathbf x^\top \mathbf A \mathbf x$ is not a positive quadratic form.

Of course, as pointed out by many, if in addition we require that $\mathbf A$ be symmetric, then all its eigenvalues are real and, moreover, $\mathbf A$ is positive definite if, and only if, all its eigenvalues are positive.