Prove that $\lambda = 0$ is an eigenvalue if and only if A is singular. I'm trying to prove that statement:
Prove that $\lambda = 0$ is an eigenvalue if and only if $A$ is singular.
I'm not sure if my proof is totally correct:
Suppose that $\lambda = 0$
if det(A) = $\lambda_1 \cdot \lambda_2 . . .\lambda_n = 0$
then A is singular.
If anyone could show me a better proof that would help.
 A: $A$ is singular $\iff x\mapsto Ax$ is not injective $\iff$ we can find $x\neq 0$ with $Ax=0\iff 0 $ is an eigenvalue of $A$.
A: Your proof is right, albeit a little unclearly written. However, you don't need the machinery of the determinant to prove this. If $\lambda=0$ is an eigenvalue of $A$, then this means there's some non-zero vector $v$ with $Av=\lambda v=0v=0$. That is, $\ker A$ is non-trivial, so $A$ is singular.
A: Given the fact that the determinant of $A$ is the product of the eigenvalues, then this is sufficient. 

Alternatively, suppose that $0$ is an eigenvalue of $A$, with corresponding non-zero eigenvector $v$. If $A$ were non-singular, then we could write
$$v = Iv = A^{-1} Av = A^{-1} 0v = 0$$
Alternatively, if $A$ is singular then $A$ must have a non-trivial null space, since it's not injective (viewed as a linear transformation). Hence $0$ is an eigenvalue.
A: I'm not sure if this quite proves it, but this is the "proof" I worked through to answer the same problem I was assigned for HW.
If λ = 0, then det(λI - A)=0 becomes det(-A)=0.
the determinant of A (regardless of the fact that it is scaled by -1 in this case) must therefore be zero, which means the matrix A is singular.
A: I know that this thread is somewhat old with an accepted answer, but I just took a course in Linear Algebra at my university and I was told that in order to prove iff instead of if you must prove that the implication goes both ways:
(1) Show that $\lambda=0\implies A$ is singular: $$\lambda=0\implies 0=|A-\lambda I|= |A-(0)I|= |A|$$
(2) Show that $A$is singular$\implies0$ is an eigenvalue of $A$: $$|A|=\Pi_{i=1}^{n}\lambda_{i}\ \land|A|=0\implies\lambda_{k} = 0\mid1\leq k\leq n$$
$$\therefore\lambda=0\iff |A|$$
