let $\dim(ker (A - \lambda I)) = 1$. why is $adj(A - \lambda I) \ne 0$

Question

Let $\lambda$ is eigenvalue of $A$ and $\dim(\ker (A - \lambda I)) = 1$.($\lambda$ has geometric multiplcity one) why is $\text{adj}(A - \lambda I) \ne 0$?

Answer 1

Let $M=A-\lambda I$; your are asking why $M$ having rank$~n-1$ implies that that adjugate matrix of $M$ is nonzero. This is because the rank is the size of the largest nonzero minor, and the adjugate of $M$ contains (up to sign) all size $n-1$ minors of $M$. The hypothesis says that at least one of those minors is nonzero, and so the adjugate of $M$ is nonzero.

For more information see answers to this question or this one.