# Find $\dim\ker(A-B)$ if the minimal polynomials of $A$ and $B$ are given

Suppose $A,B$ are $3\times 3$ matrices with minimal polynomials $x^2-4$ and $x+2$ resp. What are the possible dimensions of the kernel of $A-B$?

The possible JCFs for $A$ are $(2,2,-2), (2,-2,-2)$. The JCF of $B$ is $(-2,-2,-2)$. Thus the possible JCFs for $A-B$ are $(4,4,0),(4,0,0)$, and the possible dimensions are $1$ or $2$. (They are invariant under conjugation.)

Is this a correct sketch of solution?

• Note: $(a,b,c)$ stands for the corresp. diagonal matrix. – user419669 Aug 15 '18 at 20:31

t about the JCFs of $A$ and $B$. However your answer seems to imply that JCFs can be subtracted. The canonical form is obtained in a particular basis, and there is no reason it should be the same basis for both. However, it does work here because $B$ is a scalar matrix given that it's minimal polynomial has degree $1$. Therefore it commutes with any other matrix and will stay the same under change of bases.
• What do you mean by "$B$ is in canonical form over any basis"? Do you mean the Jordan canonical form? How do I see that it's in its JCF in any basis? That must have to do with the fact that its JCF is a scalar matrix? – user419669 Aug 16 '18 at 2:04
• @user419669 Yes, $B$ is a scalar matrix given that it's minimal polynomial has degree $1$. Therefore it commutes with any other matrix and will stay the same under change of bases. – Arnaud Mortier Aug 16 '18 at 2:36