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.

  • 1
    $\begingroup$ $v$ is an eigenvector of $A \implies Av= \lambda v$ for some $\lambda \in F$. What can you say about $v$ and $Av$ now? $\endgroup$ – Itay4 Jun 21 '17 at 8:31
  • $\begingroup$ 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. $\endgroup$ – 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 $(k_1, k_2) \in \mathbb R$.


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 $A$.



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 $$ 1 = \dim \span(\{v\}) = \dim \span(\{v, Av \}) $$


$$ \dim\span(\{v, Av\}) = 1 $$ 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 $A$.

  • $\begingroup$ 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 $Av=0$. $\endgroup$ – Michael Hoppe Jun 21 '17 at 13:12
  • $\begingroup$ Thanks, I did not notice the "iff". $\endgroup$ – mvw Jun 21 '17 at 19:25

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