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Martin Sleziak
Martin Sleziak
1:14 PM
Now I found this:
3
A: Dimension of a subspace of $M_n(\mathbb C)$.

Robert IsraelIt depends on the Jordan canonical form of $A$. For example, suppose $$ A = \pmatrix{\lambda & 1 & 0 & 0 & 0\cr 0 & \lambda & 1 & 0 & 0\cr 0 & 0 & \lambda & 0 & 0\cr 0 & 0 & 0 & \lambda & 1\cr 0 & 0 & 0 & 0 & \lambda\cr}$$ Then in order to commute with $A$, $B$ must be of the form $$ \left...

The relevant part is:
> In general, suppose $A$ has $m$ Jordan blocks of sizes $s_i$, $i= 1.. m$, and let $e(i,j) = 1$ if blocks $i$ and $j$ have the same eigenvalue, $0$ otherwise. Then the dimension is $$\sum_{i=1}^m \sum_{j=1}^m e(i,j) \min(s_i,s_j)$$
 

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