I have this question for homework that I've been trying to solve and I couldnt understand.. happy if you could help
$A$ is a square matrix with the characteristic polynomial $(t-3)^5$. It is also known that the dimension of $\text{ker}(A-3I)^2$ is $3$ and that of $\text{ker}(A-3I)^4$ is $5$.
- Is it possible that dimension of $\text{ker}(A-3I)^3$ is $3$?
- What is the geometric multiplicity of the eigenvalue $3$?
- What is the minimal polynomial of $A$?
For notational simplicity, label $A-3I$ the linear operator $T:V \rightarrow V$.
If $\dim(Ker(T^2))=3$ and if $\dim(Ker((T^3))=\dim(Ker(T(T^2)))=3$, then by rank-nullity, repeated applications of $T$ does not change the kernel. Since we know $\dim(Ker(T^4))=5$, this is impossible.
By the same reasoning, the geometric multiplicity of $3$, i.e. $\dim(Ker(T))$, must be $2$. See that if $\dim(Ker(T))=3=\dim(Ker(T^2))$ then we couldn't have $\dim(Ker(T^4))=5$, and if $\dim(Ker(T))=1$, then surely $\dim(Ker(T^2))=\dim(Ker(T(T))\neq 3$.
By Cayley-Hamilton, the minimum polynomial comes in the form $(x-3)^d$ for some $d \leq 5$. We know $T^5$ annihilates $A$, and we also know the geometric multiplicity is bounded by the algebraic multiplicity. We're given $\dim(ker(T^4))=5$, so $d \leq 4$. We have deduced $\dim(ker(T^3))=4\neq5$, so $d=4$. This proves $(x-3)^4$ is the minimum polynomial.