I'm given that $A$ is an $8 \times 8$ matrix, and then I'm given the dimensions of the kernels for $A - 2I$ raised to powers $1$ and $2$, as well as dimensions for the kernels of $A - 3I$ raised to powers $1$, $2$ and $3$.
In general terms, how do I use this information to learn about the characteristic polynomial of $A$?
I assume the Rank-Nullity theorem comes into play.


If some $(A - \lambda I)^m$ has kernel of dimension $d$, then the characteristic polynomial of $A$ is divisible by $(z-\lambda)^d$.

  • $\begingroup$ So the dimension is the degree of the associated elementary divisor? I believe you intuitively, because I know the elementary divisors generate the annihilators of the direct summands when we break down V as an F[x]-module, but it's not quite all connected in my head yet. Can you unpack this more? $\endgroup$ – BMac Jul 31 '17 at 3:00
  • $\begingroup$ Easy to see using Jordan canonical form... $\endgroup$ – Robert Israel Jul 31 '17 at 6:37
  • $\begingroup$ Because if the dimension is $d$, then the Jordan block associated with $\lambda$ is $d \times d$? Where does the power of $A - \lambda I$ come in? $\endgroup$ – BMac Jul 31 '17 at 15:52
  • $\begingroup$ The generalized eigenspace for $\lambda$ is the union of kernels of powers of $A - \lambda I$, and its dimension is the sum of the sizes of the Jordan blocks associated with $\lambda$. $\endgroup$ – Robert Israel Jul 31 '17 at 17:36

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