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Jul 18, 2012 at 8:21 comment added Geoff Robinson Finding a basis can get quite complicated. If $A$ is diagonalizable it is quite easy, but if $A$ is in Jordan normal form with Jordan blocks of size greater than $1$, determining the matrices which commute with $A$ can get quite tricky, even if $A$ has only the eigenvalue $1$ or $0$.
Jul 18, 2012 at 6:05 comment added A.S Does anyone know what is the basis for this vector space?
Jul 18, 2012 at 6:03 history answered A.S CC BY-SA 3.0