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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.

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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 $\ker (\lambda I-A)^2$.

Edit: Carl's comment.

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.

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    $\begingroup$ 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 $$Av_2' = v_1 + \lambda v_2'.$$ $\endgroup$ – Christopher Carl Heckman Apr 17 '16 at 2:38
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    $\begingroup$ And thanks for the shorter solution! I was about to post an algebraic monstrosity ... $\endgroup$ – Christopher Carl Heckman Apr 17 '16 at 2:39
  • $\begingroup$ Thanks, Carl. I incorporated your comment in the answer. $\endgroup$ – Martin Argerami Apr 17 '16 at 4:52

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