Determinant, Eigenvalues and Kernel relations Suppose the determinant of a square matrix A is 0, from what I understand, that means 0 is one of the eigenvalues of the matrix. The question is, what does it actually mean for the kernel (null space) of said matrix A?
I think it would mean that the matrix A itself is the nullspace of matrix A, and vice versa, if the determinant is not null, A is not in the nullspace.
 A: One helpful notion to think about is the eigenspace of $A$ with respect to $\lambda$. Given an matrix $A$ and a scalar $\lambda,$ define the eigenspace with respect to $\lambda$ is the following set:
$$\operatorname{Eig}(A,\lambda) = \{\text{vectors }v \text{ such that } Av=\lambda v\}.$$
You can check that this is always a vector space (that is, it's a subspace of the domain of the linear map which the matrix represents), hence the name "eigenspace".
Now there are two important observations, both easy to verify:


*

*The scalar $\lambda$ is an eigenvalue of $A$ if and only if the corresponding eigenspace $\operatorname{Eig}(A,\lambda)$ has non-zero elements.

*The kernel is just the zero eigenspace. That is, $\ker{A}=\operatorname{Eig}(A,0).$
So, in conclusion, the following are equivalent:


*

*the determinant of $A$ is zero, i.e., $\det{A}=0$;

*zero is an eigenvector of $A$;

*the zero eigenspace has non-zero elements, i.e., $\operatorname{Eig}(A,0)\neq\{0\}$;

*the dimension of the kernel is non-zero, i.e., $\dim{\ker{A}}>0.$
A: The kernel is a subset of the domain.(meaningless to say kernel is a function).
detf = O  <=> f is singular <=> ker.f is not {O}.
f(a) = O.a = O <=> a is in the kernel of f. 
