How do I find eigenspace bases in a generalized way? I have a 4x4 real matrix $A$ with three unknown eigenvalues $_{1}$, $_{2}$, $_{3}$ and the eigenspace $E_{_{3}}$ has $dim(E_{_{3}})=2$ and I have to prove that $A$ is diagonalizable but I have no idea how to do that.
I was thinking that if I could show that $_{1}$ and $_{2}$ both have 1 dimensional eigenspaces but I don't know how to do that generally. Can I just assume it? I'm not sure where to go from here.
 A: If a $4\times 4$ matrix has three distinct eigenvalues, then the sum of the dimensions of those three corresponding eigenspaces is at most $4$.  That's because their span is a subspace of $\mathbb R^4$, and eigenvectors corresponding to distinct eigenvalues are linearly independent.
In fact the eigenspaces of $\lambda_1,\lambda_2$ must be of dimension one since the eigenspace of $\lambda_3$ is known to have dimension two.  So there is no room for the other eigenspaces to be larger than dimension one.
Let's be detailed in arguing the linear independence of eigenvectors for distinct eigenvalues.  Let $u_i$ be a nonzero eigenvector for $\lambda_i$.  If there were a linear dependence relation:
$$ c_1 u_1 + c_2 u_2 + c_3 u_3 = 0 $$
with not all the coefficients $c_i$ zero, then we could multiply both sides by matrix $A$ and get a new dependence relation:
$$ c_1 \lambda_1 u_1 + c_2 \lambda_2 u_2 + c_3 \lambda_3 u_3 = 0 $$
Combining these two equations allows us to eliminate one of the eigenvectors, a procedure that can be repeated until only one term is left with a nonzero coefficient:
$$ k u_i = 0 $$
But that would imply $u_i=0$ in contradiction to our choice of nonzero eigenvectors.  So eigenspaces for distinct eigenvalues have trivial intersections, and thus the dimension of the sum of eigenspaces is the sum of their separate dimensions.
In the present circumstances that could only mean dimensions $1+1+2=4$, so we have a complete basis of eigenvectors of $A$ for vector space $\mathbb R^4$.  With respect to that basis matrix $A$ is diagonalized.
A: The total of the dimensions of the eigenspaces cannot exceed 4. You have 3 distinct eigenvalues that you have found, and one of the eigenspaces is 2-dimensional. The others must be at least 1-dimension, but they cannot exceed that because the total of the dimensions of the eigenspaces cannot exceed 4. So, the other two must be one-dimensional eigenspaces, and that means that the total of the dimensions of the eigenspaces is 4.
