Is it true that eigen vectors of matrix belong subspace that generated by vectors-"columns of matrix"?

Question

And if it is how this theorem named? If not what the counterexample?

Answer 1

Yes, it is true, unless if $v$ is an eigenvector with eigenvalue $0$ (to put it better: it may not be true then). In fact, if $A.v=\lambda v$ with $\lambda\neq0$, then$$v=\frac1\lambda A.v=A.\left(\frac1\lambda v\right)$$and therefore $v$ belongs to the range of the map $x\mapsto A.x$. But the range of that map is precisely the space spanned by the columns of the matrix.

Answer 2

An eigenvector with a nonzero eigenvalue does belong to the column space of the matrix, since $Av=\lambda v$ with $\lambda \neq 0$ implies $v = A(\lambda^{-1}v)$, where the right-hand side is obviously in the column space. But it need not be true if $\lambda=0$, as shown for example by $$ A= \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} ,\qquad v= \begin{pmatrix} 1 \\ -1 \end{pmatrix} . $$