# Prove that $v$ is an eigenvector of $A$ if and only if $\operatorname{span} (\{v, Av\})$ has dimension $1$.

If $v$ is an eigenvector of $A$ then $(A - \lambda I)v = 0$.

$$\operatorname{span} (\{v, Av\}) = av + bAv = (a + bA)v$$

If the dimension is $1$, then it wouldn't be equal to zero so I don't understand how to prove it.

• $v$ is an eigenvector of $A \implies Av= \lambda v$ for some $\lambda \in F$. What can you say about $v$ and $Av$ now? – Itay4 Jun 21 '17 at 8:31
• The expression $\span(\{v,Av\})=...$ you write in the question is conceptually wrong. The first term is a set while the second and third terms are a vector. – Jack Jun 21 '17 at 19:49

Since $v \neq 0$, the span of $\{v,Av\}$ is one-dimensional iff $\{v,Av\}$ is linearly dependent. Indeed, the only possible dimensions for for the span of $\{v,w\}$ are $0$, $1$, and $2$: $0$ if both vectors are zero, $1$ if they are linearly dependent but at least one is nonzero, and $2$ if they are linearly independent.

Now two vectors $v$ and $w$ are linearly dependent iff one is a scalar multiple of the other. For $w=Av$ and $v \neq 0$, this is equivalent to saying that $v$ is an eigenvector of $A$.

If the Span of $v$ and $Av$ has dimension $1$, then the column space is a line.

We also have that the rank of $({v_1,v_2....v_n})<n$ iff the vectors are linearly dependent.

We then let $\displaystyle \lambda=-\frac{k_1}{k_2}$.

Therefore, the vectors $v$ and $Av$ are linearly dependent, meaning that $k_1v+k_2Av=0$ for $(k_1, k_2) \in \mathbb R$.

Setting

That means that $Av$ is a scalar multiple of $v$, where the scalar is $\lambda$.

By definition of eigenvalue, $Av=\lambda v$, so $v$ must be an eigenvector of $A$.

"$$\Rightarrow$$":$$\DeclareMathOperator{span}{span}$$

For an arbitrary linear combination from $$\span(\{v, Av \})$$ one has $$c_1 v + c_2 Av = c_1 v + c_2 \lambda v = (c_1 + \lambda c_2) v \in \span(\{v\})$$ where the property of an eigenvector $$A v = \lambda v$$ was used. This means $$\span(\{v, Av \}) \subseteq \span(\{v\})$$ On the other hand, choose a linear combination with $$c_2 = 0$$, we have $$\span(\{v, Av \}) \supseteq \span(\{v\})$$ so we have $$\span(\{v, Av \}) = \span(\{v\})$$ Because $$v$$ is an eigenvector we have $$v \ne 0$$. This ensures $$1 = \dim \span(\{v\}) = \dim \span(\{v, Av \})$$

"$$\Leftarrow$$":

$$\dim\span(\{v, Av\}) = 1$$ means $$\span(\{v, Av\}) = \span(\{ u \})$$ for some $$u \in V \setminus \{ 0 \}$$.

So for an arbitrary linear combination from $$\span(\{v, Av \})$$ there must exist a scalar $$c$$ which allows us to write $$c_1 v + c_2 A v = c u$$ We pick a linear combination with $$c_1 \ne 0$$ and $$c_2 = 0$$ and get $$c_1 v = c u \iff \\ v = \frac{c}{c_1} u = \alpha u$$ so $$v$$ is a scalar multiple of $$u$$.

If we had $$v = 0$$ then the linear combinations from $$\span(\{v, Av \})$$ would reduce to $$c_1 v + c_2 Av = c_1 0 + c_2 A 0 = 0$$ because $$A$$ is linear and the span would consist only of the null vector, which would contradict $$\dim \span(\{v, Av \}) = 1$$. So we have $$v \ne 0$$. Further this implies $$\alpha \ne 0$$ because we had $$u \ne 0$$.

We pick a linear combination with $$c_1 = 0$$ and $$c_2 \ne 0$$ and get $$c_2 Av = c u \iff \\ Av = \frac{c}{c_2} u = \frac{c}{c_2 \alpha} v = \lambda v$$ so $$v$$ is an eigenvector of $$A$$.

• I think you have to investigate a bit more on the "$dim=1\Rightarrow v$ is an eigenvector"-direction: one has to discuss the cases $v=0$ and $Av=0$. – Michael Hoppe Jun 21 '17 at 13:12
• Thanks, I did not notice the "iff". – mvw Jun 21 '17 at 19:25