We can conclude second option with an example,
$\begin{bmatrix} 2 & -2& 1\\ -1& 3& -1\\ 2 & -4 & 3 \end{bmatrix}$
Skipping the calculation step of eigen values. Eigenvalues are $\lambda1$ = 1,$\lambda2$ = 1,$\lambda3$ = 6,
Eigen Vectors corresponding to eigen value, $\lambda1$ = 1,
Eigen vector you will get in terms of two linearly independent variable, I have taken k1 and k2 and the vector will be,
(2k1-k2,k1,k2) i.e k1 ( 2,1,0)+k2(-1,0,1)
Now you will see that these two vectors $\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$ & $\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ corresponding to one eigen value,$\lambda1 = 1 $.
Both will satisfy the characteristic equation, $\left ( A-\lambda I \right ) X = 0$
Checking for $\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$ ,
$\begin{bmatrix} 2 -1& -2& 1\\ -1 & 3-1& -1\\ 2 & -4 & 3-1 \end{bmatrix}$ * $\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$
= $\begin{bmatrix} 1& -2& 1\\ -1 & 2& -1\\ 2 & -4 & 2\end{bmatrix}$ * $\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$
= $\begin{bmatrix} 2 * 1 + (-2) * 1+1*0\\ (-1)*2 + 2*1 + (-1)*0\\ 2*2 + (-4)*1 + 2*0 \end{bmatrix}$
= $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
Checking for $\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ ,
$\begin{bmatrix} 2 -1& -2& 1\\ -1 & 3-1& -1\\ 2 & -4 & 3-1 \end{bmatrix}$ * $\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$
= $\begin{bmatrix} 1& -2& 1\\ -1 & 2& -1\\ 2 & -4 & 2\end{bmatrix}$ * $\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$
= $\begin{bmatrix} 1*(-1) + (-2)*0+ 1*1\\ (-1)* (-1) + 2 *0 + (-1 )*1\\ 2* (-1) + (-4) * 0 + 2 *1\end{bmatrix}$
= $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
Moreover these vectors are linearly Independent,
C1*$\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}$ +C2* $\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$ =0
The above equation are only possible for C1 = C2 = 0, So both vectors are linearly independent as well.
So its not always the case that if V1 and V2 are two linearly independent eigen vectors then they correspond to distinct eigen values.
