If eigenvalues are positive, is the matrix positive definite?

If the matrix is positive definite, then all its eigenvalues are strictly positive.

Is the converse also true?
That is, if the eigenvalues are strictly positive, then matrix is positive definite?
Can you give example of $2 \times 2$ matrix with $2$ positive eigenvalues but is not 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.