Singularity positive semidefinite 
*

*The determinant of a matrix equals the product of its eigenvalues.

*A positive semidefinite matrix is a symmetric matrix with only nonnegative eigenvalues.

*A positive definite matrix is a symmetric matrix with only positive eigenvalues.


Combining (1) and (3) yields that a positive definite matrix is always nonsingular since its determinant never becomes zero.
Is is true that for a positive semidefinite matrix at least one of its eigenvalues equals zero and thus its determinant always equals zero => a positive semidefinite matrix is always singular?
You would say that specifically having a positive semidefinite matrix instead of a positive definite matrix implies that at least one of the eigenvalues equals zero.
Is this correct?
 A: I see there is a bit of confusion, so I'll try to explain better.
A matrix $A$ is positive semidefinite (p.sd.) if it is symmetric and all its eigenvalues are $\ge 0$.
A matrix $A$ is positive definite (p.d.) if it is symmetric and all its eigenvalues are $>0$.
This means that every p.d. matrix is also a p.sd. matrix. The set of positive semidefinite matrices contains the set of positive definite matrices, but there are some p.sd. matrices that are not p.d., like the matrix that has all elements equal to zero.
As an analogy, you can think about the real and integer numbers: all integer numbers are real numbers, because the set of real numbers contains the set of integer numbers, but there are real numbers, such as $\sqrt 2$ that are not integer.
The most common example is the identity matrix $I$: all its eigenvalues are $1>0$, so it is a p.d. matrix, but $1\ge 0$ so it is also a p.sd. matrix. This means that $I$ is a p.sd. matrix and it is invertible.
A: If a matrix $M$ is Positive definite then for all non-zero $x$, $x^TMx \gt 0$.
If a matrix $M$ is Positive semidefinite then for all non-zero $x$, $x^TMx \ge 0$.
So, every positive definite matrix is positive semidefinite, but not vice versa. 
If there is a matrix $S$ which is positive semidefinite but not positive definite then at least one of its eigen values is zero, hence it is a singular matrix. For proof, you can use schur factorization to diagonalize the matrix $S$ such that $S = QDQ^T$ where $D$ is a real diagonal matrix formed using the eigen values of $S$, and $Q$ is a real (by construction) unitary (hence full-rank) matrix.
