Does eigenvectors of a matrix change during matrix operations? If I have a matrix $A$ with two eigenvectors $x$ and $y$. 
What will be the eigenvectors of 
$$A^2 - 3A + 4I ?$$
I know that if we take powers of $A$ then the eigenvectors remain unchanged. But I am not quite sure about the above matrix .
 A: If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$,
$Av = \lambda v, \tag{1}$
and $p(A)$ is a polynomial in $A$,
$p(A) = \sum_0^n p_i A^i, \tag{2}$
then $p(\lambda)$ is an eigenvalue of $p(A)$, also with eigenvector $v$, since
$p(A)v = (\sum_0^n p_i A^i)v = \sum_0^n p_i A^iv = \sum_0^n p_i \lambda^i v = (\sum_0^n p_i \lambda^i)v = p(\lambda)v; \tag{3}$
thus if 
$p(A) = A^2 -3A + 4I, \tag{4}$
then
$p(A)v = (\lambda^2 - 3\lambda + 4)v. \tag{5}$
If $x$ and $y$ are two eigenvectors of $A$, the above indicates they will also be eigenvectors of $p(A)$.
A: Evaluation of polynomials of a matrix is compatible with change of basis, in other words taking a polynomial of $A$ and then doing a change of basis has the same effect as doing the change of basis directly on$~A$ and then evaluating the same polynomial of the resulting matrix. You can check this by explicit computation using the formula for change of basis (the basic case is where the polynomial is just taking some $k$-th power), but it is more insightful to say that evaluating a polynomial of a linear operator (defined by $A$  relative to a given basis) is well defined, regardless of the coordinates used to describe such an operator, and changing bases is just a switch between two points of view of the same situation.
Then if $A$ is diagonalisable, you can do a change of basis to a situation in which the matrix is actually diagonal, and your question becomes whether a polynomial of a diagonal matrix is still a diagonal matrix, and what can be said of their diagonal entries. Since the set of diagonal matrices is closed under addition, multiplication, and scalar multiplication, the first question has an affirmative answer, and it is clear that evaluating a polynomial $P$ replaces each diagonal entry $a$ by the evaluation $P[a]$.
A: If we have $Ax=\lambda x$, then this implies $$(A^2-3A+4I)x=A^2x-3Ax+4x=\lambda^2x-3\lambda x+4x=(\lambda^2-3\lambda+4)x$$ hence $x$ is an eigenvector of $A^2-3A+4I$ to the eigenvalue $\lambda^2-3\lambda+4$
