# Let $A$ be a $2\times 2$ matrix, s.t. $p(x)=(x-\lambda)^2$ is its characteristic polynomial.

Let $\dim\ker(\lambda I-A)=1$, and $v_1\in\ker(\lambda I-A)$; $v_1\neq0$.

Also, we know that, then, $\dim\ker(\lambda I-A)^2=2$. Take $v_2\in \ker(\lambda I-A)^2\setminus\ker(\lambda I-A)$; $v_n\neq 0$.

Why can one say that $Av_2=v_1+\lambda v_2$?

I know that $Av_2\in\ker(\lambda I-A)^2$, then there are $\alpha$ and $\beta$ s.t. $$Av_2=\alpha v_1+\beta v_2$$ bu how can one conclude that $\alpha=1$ and $\beta=\lambda$?

I'm trying to understand Jordan Normal Form. Because we know that in this case $$J=\left(\begin{array}{cc} \lambda & 1\\ 0 & \lambda\end{array}\right),$$ in a way that $\{v_1,v_2\}$ is its Jordan basis.

You have $$0=(\lambda I-A)^2v_2=(\lambda I-A)\,(\lambda I-A)v_2.$$ This shows that $(\lambda I-A)v_2\in\ker(\lambda I-A)$. Since $\dim\ker(\lambda I-A)=1$, any vector in it is a scalar multiple of $v_1$. Thus $$(\lambda I-A)v_2=\alpha\,v_1$$ for some scalar $\alpha$. This we can rewrite as $$Av_2= - \alpha v_1 + \lambda v_2.$$ Note that the constant in front of $v_1$ cannot be avoided if you are taking $v_1$ as any element of $\ker(\lambda I-A)$ and $v_2$ as any element of $\ker (\lambda I-A)^2$.
Now, if we allow ourselves to choose $v_2$, then we can replace the $v_2$ above by $v_2/(-\alpha)$, and then the formula becomes $$Av_2=v_1+\lambda v_2.$$ Note that $\alpha\ne0$, because otherwise $v_2$ would be an eigenvector for $\lambda$, which it is not by choice.
• As to why $\alpha=1$, first of all note that if $\alpha=0$, then $v_2$ would be an eigenvector itself (which it isn't). Hence, you can divide both sides of the equation by $-\alpha$ to get $$A \cdot {v_2 \over -\alpha} = v_1 + \lambda \cdot {v_2 \over -\alpha},$$ so if we use $\displaystyle v_2'= {v_2 \over -\alpha}$ instead of $v_2$, then $$Av_2' = v_1 + \lambda v_2'.$$ – Christopher Carl Heckman Apr 17 '16 at 2:38