Why does $(A- \lambda I)^2 =0$ if A has two repeated eigenvalues?

Question

This statement appears in my textbook as part of an introduction of the method for finding the Jordan form of a $2 \times 2$ matrix. I understand what it says but I'd really like to know where is it coming from or what is the proof of it.

Thanks a lot!

Answer 1

If $A$ is a $2\times 2$ matrix that has and eigenvalue $\lambda$ repeated twice, then the characteristic polynomial of $A$ is $(x-\lambda)^2$, and so by the Cayley-Hamilton theorem, which says that a matrix annihiliates its characteristic polynomial, you get $(A-\lambda I)^2=0$.

Answer 2

If $A$ has two repeated eigenvalues $\lambda$, then its Jordan form is $$ \begin{pmatrix} \lambda & * \\ 0 & \lambda \end{pmatrix} $$ in other words, $A$ is conjugate to a matrix of the above form. Since conjugate matrices have the same characteristic polynomial, and the characteristic polynomial of the above matrix is $(x-\lambda)^2$. By Cayley-Hamilton, the claim follows.