Are all strictly positive semi-definite matrices singular?

Question

If I have some matrix A with an eigenvalue of 0, what makes this matrix singular? and I am assuming All positive definite matrices are non singular so all strictly positive-semidefinite matrices would be singular? How do eigenvalues relate to singularity is my underlying question.

Answer 1

If $0$ is an eigenvalue of $A$ then $A\mathbf v=0\mathbf v=\mathbf 0$ for some non-zero eigenvector $\mathbf v$, but we also have $A\mathbf 0=\mathbf 0$. Since $A\mathbf v=A\mathbf 0$ but $\mathbf v\neq \mathbf 0$, the map defined by $A$ is not one-to-one and hence not invertible. In other words, $A$ is singular.

Answer 2

Your question is a bit strange. A square matrix $A$ over a commutative ring is said to be singular if $Ax=0$ for some nonzero vector $x$. Note that this notion is implied by but not equivalent to the invertability of $A$, although the two notions are equivalent when the ring is a field.

Anyway, if $A$ is a matrix over some field and it has a zero eigenvalue, then $Ax=0$ for some nonzero vector $x$. Hence $A$ is singular. Unless you are using some weird definition of singularity, the singularity of $A$ is an immediate consequence of the existence of a zero eigenvalue.