How to prove matrix $A$ is invertible $\iff$ $\lambda=0$ is not an eigenvalue of $A$?

Question

So we know that $A$ is not invertible $\iff$ det$(A)=0\iff \lambda=0$ is an eigenvalue of $A$. But the negation doesn't equal to the title statement. How would you prove the title question?

Answer 1

Straight from the definition... $\lambda = 0$ is an eigenvalue if and only if there is a $v \neq 0$ such that $Av = 0$. But this happens if and only if $\ker A \neq \{ 0 \}$, which happens if and only if $A$ is not invertible.

Geometrically, an eigenvalue of zero means that the transformation $A$ compresses the line spanned by $v$ to a point, i.e., applying $A$ reduces dimensions. Thus it cannot have full rank.

Answer 2

• If $det(A)=0$ then $0$ is an eigenvalue.

• If $0$ is an eigenvalue then $det(A)=0$. Recall that for a given conditional its contrapositive is always true.