proof that a matrix $M$ is positive definite iff all its eigenvalues are strictly positive. By definition, an $n\times n$ matrix $M$ is positive definite iff
$$\forall z\in\mathbb R^n:z^TAz>0$$
We want to show that this is satisfied iff for all eigenvalues $\lambda _i $ of $M$, we have
$$\lambda_i >0$$
We can show this by decomposing $M$ into a diagonal matrix: $$M=P^{-1}DP$$
where $P$ is the matrix built up by the eigenvectors of $M$, and where $D$ is the diagonal matrix with the eigenvalues on the main diagonal.
Then if we dfine $y=Pz$, we can rewrite the criterion as:
$$z^TP^{-1}Dy>0$$
However, in the text I found on wikipedia, they go one step further, and rewrite it as 
$$\forall y\in \mathbb R^n: y^TDy>0$$
From that condition it is easy to see that all eigenvalues must be strictly positive, but I don't know how this conclusion is reached since $y^T=z^TP^T\neq z^TP^{-1}$
ps. here is the source:


Edit: I now understand that $z^TP^T=z^TP^{-1}$ is based on the assumption that $M$ is symmetric and therefore $P$ is orthogonal. However, I am wondering, can we still say that a non-symmetric $M$ is positive definite iff all its eigenvalues are strictly positive?
 A: Your source talks about matrices over the complex numbers and $M$ is supposed to be Hermitian, that is, equal to its “conjugate transpose”.
I'll denote by $M^*$ the conjugate transpose of the (generic) matrix $M$. A couple of definitions; let $M$ be a square matrix:


*

*$M$ is Hermitian when $M=M^*$

*$M$ is normal when $MM^*=M^*M$
The source also talks about the spectral theorem, which says that every normal matrix $M$ can be written as
$$
M=\sum_{k=1}^r \lambda_k P_k\tag{*}
$$
where $\lambda_1,\dots,\lambda_r$ are the distinct eigenvalues of $M$ and $P_1,\dots,P_r$ are the projections matrices to the eigenspaces; in particular they satisfy $P_k^2=P_k$ and $P_k^*=P_k$. Moreover $P_kP_l$ is the null matrix when $k\ne l$.
Clearly any Hermitian matrix is normal and it's easy to deduce that if $M$ is Hermitian then its eigenvalues are real: indeed, using its spectral decomposition (*), we have, from $M=M^*$,
$$
\sum_{k=1}^r \lambda_kP_k=\sum_{k=1}^r \lambda_k^*P_k
$$
and upon multiplying this by $P_l$ we get $\lambda_l=\lambda_l^*$.
The statement that any Hermitian matrix has real eigenvalues can be proved also without the spectral theorem, by directly using the definition: let $v$ be an eigenvector relative to the eigenvalue $\lambda$; then
$$
\lambda(v^*v)=v^*(\lambda v)=v^*Av=v^*A^*v=(Av)^*v=(\lambda v)^*v=\lambda^*v^*v
$$
Since $v^*v\ne0$, we get $\lambda=\lambda^*$.
Another important fact about normal matrices is that they can be diagonalized with a unitary matrix; this is actually an equivalent condition.

A square matrix $M$ is normal if and only if there exists a unitary matrix $U$ (that is, $U^{-1}=U^*$) such that $D=U^{-1}MU$ is diagonal.

Now, suppose $M$ is (Hermitian) and positive definite, that is, for $v\ne0$, $v^*Mv>0$. Consider $e_k$, the $k$-th vector of the canonical basis and consider $v_k=Ue_k$; then
$$
0<v_k^*Mv_k=e_k^*U^*UDU^*Ue_k=e_k^*De_k
$$
and, clearly, $e_k^*De_k$ is the entry at place $(k,k)$ in $D$. Since $D$ has on its diagonal the eigenvalues of $M$, we are done.
Conversely, if $D$ has all its diagonal entries positive, then, for every $v\ne0$ we have
$$
v^*Mv=(Uv)^*D(Uv)>0
$$
because clearly $D$ is positive definite.
