İs it true if $0$ is an eigenvalue of a matrix $A$ then $\det{A}=0$?

Question

İs it true if $0$ is an eigenvalue of a matrix $A$ then $\det{(A)}= 0$ ?

Is there any explanation?

Answer 1

$0$ is an eigenvalue iff $\exists x\neq 0\, \, Ax=0\cdot x=0$. This means that $\ker{A}\neq\{0\}$. The linear application represented by $A$ is not injective and $\det{A}=0$

Answer 2

Because $|A-\lambda E|=0$ for $\lambda=0$.

Answer 3

When $A$ has eigenvalue $0$, that means there exists a vector $X$ for which $AX=0$.

This is a homogenous linear system, which can only have a non-trivial solution when $\det(A)=0$, because otherwise there would be a solution $A^{-1}AX=A^{-1}\cdot0$ and that implies that $X=0$

But because $0$ is an eigenvalue, we know that there exists a non-zero vector $X$ for which $AX=0$ and thus $\det(A)=0$