All,
I encounter the following statement in one note:
"$v$ is an eigenvector of matrix $A$ " is equivalent to : rank$(A,v) \leqslant$ 2"
however, I cannot figure that how the above equivalency can hold.
I try to think from the "rank-nullity theorem", but what I can have is:
Row number of matrix $(A,v) = n+1 =$ dim$(\text{kernel}(A,v)) + \text{rank}((A,v))$,
then that means what we seek is to verify : dim$(\text{kernel}(A,v)) >= n-1$
In fact, I can construct the following relationship from the definition of the eigenvalue, i.e. assume $\lambda$ is the associated eigenvalue for $v$, then:
$$Av-\lambda v =0 \iff \begin{bmatrix} A, v \\ \end{bmatrix} \begin{bmatrix} v \\ \lambda \\ \end{bmatrix} =0 $$
But then?
As stated, the assertion is clearly false. Take $A $ to be any matrix with rankmat least three, and $v $ an eigenvector for $A $. For instance, let $A=I_3$ and let $v \in\mathbb R^3$ be any nonzero vector. Then $Av=v $, and $$\text {Rank}(A,v)=3. $$