why does $A−λ_iI$ lose rank?

Question

why does $A−λ_iI$ lose rank?

In other words, why does rank(A) is larger than rank($A−λ_iI$)?

Answer 1

This is only true if $0$ is not an eigenvalue of $A$, which is to say that $A$ is invertible.

By definition, $\lambda_i$ is an eigenvalue exactly when $A-\lambda_i I$ is not invertible, which is to say that $A-\lambda_i I$ is not of full tank. If $A$ is invertible, this means precisely that $A-\lambda I$ has lower rank precisely when $\lambda$ is an eigenvalue.