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Let A be a square matrix and $v,w$ be non-zero vectors satisfying $(A-\lambda I)v=0$ and $(A-\lambda I)w=v$ for some $\lambda \in \mathbb{C}$. Show that $v$ an eigenvector and $w$ a generalized eigenvector are linearly independent.

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    Well, *two* vectors are *not* linearly independent, only if one is the *the same* vector as the other, rescaled by a coefficient... – 2017-01-23
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    Unfortunately, $Aw\ne\lambda w$. – 2017-01-23
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    You could consider $(A-\lambda I)(a_1v+a_2w)$. – 2017-01-23
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    @SergeiGolovan thanks for pointing that out. – 2017-01-23
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    @SergeiGolovan I fail to understand your objection. If $v$ and $w$ were non-zero and linearly dependent, then $w=av$ and $v=(A-\lambda I)w=(A-\lambda I)av=0$. – 2017-01-23
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    @Anonymous: You're correct. But I don't see anything similar to your reasoning in the original question. It's just stated that $Aw=\lambda w$ out of the blue. – 2017-01-23
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    @Anonymous I mistook the generalized eigenvector for an eigenvector. – 2017-01-23
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    @SergeiGolovan I must admit I'm confused, but I'll probably just drop it after stating: if someone says "prove that two non-zero vectors are linearly independent" he's clearly saying "prove that they are *not* the same vector, rescaled". I think this is immediate if one is in the nucleus of *any* matrix and the other isn't... – 2017-01-23

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I have found an answer to my own question: First note that $Aw-\lambda w =v \to Aw =v+\lambda w$ and $Av=\lambda v$. \begin{align} (A-\lambda I)(a_1v+a_2w)&=0 \\ a_1Av+a_2Aw -(a_1\lambda v+a_2 \lambda w)&=0 \\ a_1\lambda v+a_2(v+\lambda w) -(a_1\lambda v+a_2 \lambda w)&=0 \\ a_1\lambda v -a_1\lambda v+a_2\lambda w -a_2 \lambda w + a_2v&=0 \\ a_2v&=0 \end{align} As $v$ is non-zero $a_2=0$. Subbing in $a_1v+(0)w=0$ we get $a_1v=0$ so $a_1=0$. As both $a_1=a_2=0$ the vectors $v$ and $w$ are linearly independent.

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    @SergeiGolovan I took you advice is my solution correct? – 2017-01-23