[Math] Why is $\det(A – \lambda I)$ zero

Question

I was reading over some notes of mine from the Linear Algebra class I took a while ago in order to prepare myself for Computer Graphics course I'll be taking this Fall and I ran across a sentence that I can't quite grasp. It states that

Since we have a non-zero vector $x$ in the nullspace of $A – \lambda I$, then $\det(A – \lambda I) = 0$.

From reading Wikipedia page on determinants I got that

… the system has a unique solution exactly when the determinant is nonzero; when the determinant is zero there are either no solutions or many solutions.

So if we have vector $x$ in the nullspace of $A-\lambda I$, then we have either 0 solutions or many solutions. But I still don't understand why $\det(A – \lambda I)$ has to be equal zero?

Answer 1

If $x \ne 0$ is in the null space of $A - \lambda I$, i.e., $(A - \lambda I)x = 0$, that means that $A$ is singular (noninvertible), which is exactly the same as saying that its determinant is zero.