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?
 A: 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.
A: The converse will be true if the matrix is diagonalizable.  That's why the counter-example given by Ronaldo (edited by KennyTM) is not diagonalizable.  If $B$ is diagonalizable, then $B=P^TAP$ and $y^TBy=y^TP^TAPy=x^TAx=\sum d_ix_i^2$ (where $P$ is an orthogonal matrix whose transpose is its own inverse) as in the answers by Soarer.
Spectral theorem for non-diagonalizable matrix gives rise nilpotent matrices, i.e.
$$A=\sum_i (\lambda_i P_i + N_i)$$ where $P_iP_j = \delta_{ij} P_j$, $N_iP_i=N_i$ etc. ($P_i$'s are projection matrices and $N_i$'s are nilpotent matrices corresponding to eigen-values $\lambda_i$).
A: 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.
A: 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.  
A: True. If you consider a diagonal matrix $A = \mathrm{diag}(d_1,\cdots,d_n)$ where each diagonal entry $d_i$ is positive, then clearly it is positive definite, since $x^TAx = \sum d_ix_i^2 > 0$ unless $x = 0$ ($x_i$ are the components of vector $x$.)
Now apply spectral theorem for symmetric matrix to reduce to the diagonal case.
A: In general, the converse is not true but if we assume our matrix M to be Hermitian then the converse becomes also true. 
By assuming our matrix to be Hermitian we have that it is unitary diagonalizable which is used to prove that it is Positive definite.
A: For a matrix to be positive definite:
1) it must be symmetric
2) all eigenvalues must be positive
3) it must be non singular
4) all determinants (from the top left down the diagonal to the bottom right - not jut the one determinant for the whole matrix) must be positive.
If a 2x2 positive definite matrix is plotted it should look like a bowl. If the matrix is singular then it's a trough which follows the vector which takes the matrix to zero. If the sum of the cross terms is bigger than the trace then the plot is a saddle at zero. 
