$Av=\lambda v \implies$ $Av$ is parallel to $v$.
If one eigenvalue is 0, then the determinant is 0 and the matrix is singular.
What is the meaning of a matrix eigenvalue to be 0?
Is it like $Av$ will disappear from the space?
Any physical meaning or analogy to help understand it?
P.s: I am not looking for equations.
First consider the case of a matrix $A$ in $\mathbb{R}^{n\times n}$. Then $x\mapsto Ax$ is a linear transformation on $\mathbb{R}^{n}$. According to the theory of linear algebra, $A$ can be finally decomposed into dilations in some direction and rotations in some other directions.
A dilation corresponds to a real eigenvalue and an eigenvector, which are, in your case, $\lambda$ and $v$. If $\lambda=0$, then for all vectors in the direction of $v$ we have $A(cv)=cAv=c\lambda v=0$, i.e., the direction $v$ disappears under the action of $A$.
A rotations corresponds to a nonzero complex eigenvalue, which I will not go into details since you seem to be concerned only about the eigenvalue $0$.
An example might help you understand. Let $$A=\left(\begin{matrix}1&2\\0&0\end{matrix}\right)$$ It has two eigenvectors: $(2,-1)^T$ with e.v. $0$, and $(1,0)^T$ with e.v. $1$. Then you set up a new frame generated by these two vectors, and decompose any other vector in these two directions. Then the action of $A$ as a linear map can be said as follows:
(1) dilate a vector in the direction of $(1,0)^T$ by a factor of $1$. (2) dilate a vector in the direction of $(2,-1)^T$ by a factor of $0$, i.e., ''disappear''.