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

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!

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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 $$(A-\lambda I)^2=0$$.
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.