# Generalised eigenvalue is eigenvalue if it is in the field

I would like to prove the following assertion:

Let $\mathscr{F}$ be a field and $\mathscr{\phi}$ be an $\mathscr{F}$-linear endomorphism of a finite dimensional $\mathscr{F}$-vector space $\mathscr{V}$. Then, if $\lambda \in \mathscr{F}$ is a generalised eigenvalue, $\lambda$ is an eigenvalue.

Translating the requirement, we see that what we want to show is the following:

Let $\mathbf{v} \neq 0$ satisfy the equation $$(\phi - \lambda \operatorname{id}_{\mathscr{V}})^k\mathbf{v} = \mathbf{0}$$ for some natural $k \ge 1$. We want to establish that $$(\phi - \lambda \operatorname{id}_{\mathscr{V}})\mathbf{v} = \mathbf{0}$$

My first attempt for this problem was proof by contradiction. Suppose that:

$$(\phi - \lambda \operatorname{id}_{\mathscr{V}})\mathbf{v} \neq \mathbf{0}$$

I was thinkinh of multiplying the above with $$\phi - \lambda \operatorname{id}_{\mathscr{V}}$$ $k$-times in hopes that I will get $$(\phi - \lambda \operatorname{id}_{\mathscr{V}})^k\mathbf{v} \neq\mathbf{0}.$$

In case this strategy is correct, would someone be willing to continue? If this is incorrect, would someone be willing to propose a correct strategy?

Thank you very much in advance! All help is greatly appreciated!

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Let $\mathbf{v} \neq \mathbf{0}$ be a generalised eigenvector for the generalised eigenvalue $\lambda$, and let $k \geq 1$ be the least natural number such that $(\phi - \lambda \operatorname{id}_{\mathscr{V}})^{k}\mathbf{v} = \mathbf{0}$. Let $\mathbf{w} = (\phi - \lambda \operatorname{id}_{\mathscr{V}})^{k-1}\mathbf{v}$. By our assumption on $k$, $\mathbf{w} \neq \mathbf{0}$; what is $(\phi - \lambda \operatorname{id}_{\mathscr{V}})\mathbf{w}$?

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Thank you very much for your help, Branimir! – Orest Xherija Apr 12 '13 at 2:54

Your problem may be that you've incorrectly stated what you need to prove. When you assume

$$(\phi - \lambda \text{id}_{\mathscr{V}})^k \mathbf{v} = 0$$

then your goal is not to show that

$$(\phi - \lambda \text{id}_{\mathscr{V}}) \mathbf{v} = 0$$

but instead that there exists a nonzero vector $\mathbf{w}$ (which might be $\mathbf{v}$ or it might be something else entirely) satisfying

$$(\phi - \lambda \text{id}_{\mathscr{V}}) \mathbf{w} = 0$$

Once you have straight what you need to prove, you can then continue as in Branimir's answer.

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I see my mistake! Thank you very much for pointing this out! – Orest Xherija Apr 12 '13 at 2:54