Two questions on eigenvalue theory 
$6.)$ The vector $v = (1,−2,1)$ is an eigenvector of the real matrix 
$$A =\begin{bmatrix}
a &b& c\\
4& 1& 2\\
d& e& f
\end{bmatrix}$$
Then $v$ is an eigenvector of $A^3$ with corresponding eigenvalue
(A) $64$
(B) $−8$
(C) $1$
(D) $8$
(E) $−1$
Answer: B
$7.)$ Suppose that $A$ is a $3\times 3$ real matrix with $k$ distinct real eigenvalues and $l$ distinct complex (i.e., non-real) eigenvalues. Consider the following statements: 
(i) $k = 2,\ l = 0$ 
(ii) $k = 0,\ l = 3$
(iii) $k = 1,\ l = 2$
(iv) $k = 2,\ l = 1$ 
How many of the above statements are possible?
(A) None
(B) $1$
(C) $2$
(D) $3$
(E) All of them
Answer: C

For $6.)$ I know $A$ times the eigenvalue gives $\lambda$ times the eigenvector but the matrix has letters in it and I cant seem to get the right and when I multiply $A$ by the eigenvalue.
For $7.)$ do you multiply the top by the conjugate?
 A: 6) Since $Av = \lambda v $ then 
$A^3v = A^2 Av = A^2 (\lambda v) = .... = \lambda^3 v $
But then 
$4 (1) + 1 (-2) + 2 (1) = 4 = -2 \lambda$ which implies $\lambda = -2$ hence $\lambda^3 = -8$.
7) take $det (A - \lambda I) = 0$ which in a $3 \times 3$ matrix is a 3rd degree polynomial of the form $p (\lambda) = a_3 \lambda^3 + a_2 \lambda^2 + a_1 \lambda + a_0$
Over any algebraically closed field this polynomial has at least one root. Specifically over the Complex field it has 3 roots. If any of those roots happen to be complex numbers (not real), then we know from the Complex Conjugate Pair Theorem that the conjugate of that root is also a root. Thus $l $ is even (where 0 is an even number).  
So we can tease out the answer (even though we don't know the field). (i) can be true since we could have a repeat real root and a separate distinct root. (ii) cannot be true since there cannot be an odd number of roots. (iii) works fine, 1 real 2 complex roots. (iv) fails since there is a lone complex root without its conjugate.
