Eigenspaces and jordan normal form I have a question here regarding the jordan normal form of two matrices where the eigenspace is one is contained in the other.
Let $A,B$ be two $n \times n$ matrices s.t $AB=BA$. I firstly proved that the eigenspace of A is preserved by B, getting the result that $BV(\lambda) \subset V(\lambda)$ where $V(\lambda) = \{ x | Ax = \lambda x\}$. 
I assume $A$ is diagonalisable and have to prove that there exists an invertible matrix $T$ s.t both $TBT^{-1}$ and $TAT^{-1}$ are in jordan normal form. Since $A$ is diagonalisable, it is similar to a diagonal matrix. So this means that there exists invertible $T$ such that $TAT^{-1}$ is diagonal, which I believe is also its Jordan normal form just with jordan blocks of size $1$. 
Since the eigenspaces of $B$ is contained in $A$, then an eigenvalue of $B$ is necessarily an eigenvalue of $A$, so it will have the same eigenvalues, but not necessarily as many jordan blocks. How can i further go onto prove that $TBT^{-1}$ is in Jordan normal form?

This won't quite work: e.g. you might have $A = I$, in which case your $T$ could be any invertible matrix, and there's no reason to think $T B T^{-1}$ will be in Jordan normal form.  What you want to do is choose $T$ on each eigenspace of $A$ to put into Jordan form the restriction of $B$ to that eigenspace. 
