Is there a eigenvalue equal to 0 if determinant is equal to 0? According to theorem the multiplication of all eigenvalues is equal to the determinant, so if one of them equals 0 the determinant is always 0. But is it true for the opposite statement? If determinant is equal to 0 is there for sure an eigenvalue equal to 0?
 A: If a square matrix $A$ has zero determinant, this implies that $A$ is not injective, i.e. the kernel is nonempty. So, there exists $v \neq 0$ such that $Av = 0 = 0\cdot v$. By definition, $v$ is an eigenvector for $A$ corresponding to eigenvalue $0$.
A: Maybe you've heard of the characteristic polynomial. For a size$~n$ square matrix $A$ it is defined as $\det(XI_n-A)$, it is a monic polynomial of degree$~n$, and it has the property that $\lambda$ is an eigenvalue if and only if $\lambda$ is a root of the characteristic polynomial, in other words if $\det(\lambda I_n-A)=0$. Setting $\lambda=0$ in this statement, it says that $0$ is an eigenvalue if and only if $\det(-A)=0$. It is not very hard to see that this is equivalent to $\det(A)=0$.
A: Yes, the determinant of a matrix with real/complex entries is the product of its (complex) eigenvalues, so it has a matrix has a $0$ eigenvalue if and only if its determinant is $0$.
If you care about matrices with entries in a general field $F$, then as Clement points out the determinant will be the product of the eigenvalues which lie in the algebraic closure of $F$, and so once again a matrix has a $0$ eigenvalue if and only if its determinant is $0$.
A: Fun, but overkill solution:
Zero determinant implies one of the singular values must be zero. By Weyl's Inequality, one of the eigenvalues must be zero.
