Characteristic polynomial of a linear endomorphism of dimension $n$.

Question

So, if $T: V \rightarrow V$ and I suppose that $T^2-3T+2I=0$ and that the $\mathrm{rank}(I-T)=k$, what would be the characteristic polynomial of $T$?

I know from previous questions that the eigenvalues of $T$ are $1$ and $2$ and that $T$ is invertible with $T^{-1}=\frac{3}{2} I - \frac{1}{2} T$.

I also know that $T$ is diagonalizable and $\mathrm{rank}(T-I)=n-\dim(\mathrm{Ker}(2I-T))$ where $n=\dim(T)$.

I am just having a bit of trouble trying to figure out the characteristic polynomial of $T$.

What I have worked out so far is that when there is a basis $B$ such that $\det(xI-[T]_{B \leftarrow B})=(x-1)^a(x-2)^b$ since those are the only eigenvalues but since the dimension is $n$. I'm not exactly sure how I would find $a$ and $b$. Any help would be appreciated.

Answer 1

The eigenvalues of $T$ belong to the set of roots of the annihilator polynomial $P(x)=x^2-3x+2$ of $T$ which is the set $\{2,1\}$ and since $P$ has simple roots so $T$ is diagonalizable and then the geometric and algebraic multiplicities of the eigenvalues are equal. Since $rank(I-T)=k$ then $\dim\ker(T-I)=n-k$ so $1$ is an eigenvalue of $T$ of multiplicity $n-k$ and then $2$ is an eigenvalue of multiplicity $k$ and we get

$$\chi_T(x)=(x-1)^{n-k}(x-2)^k$$

Answer 2

A vital point to remark is that the given annihilator polynomial $X^3-3X+2$ is split with simple roots $1,2$, which ensures that $T$ is diagonalisable with eigenvalues among $\{1,2\}$ (this is a fundamental theorem). The rank information, which will give the dimension of the eigenspace for $\lambda=1$, therefore also gives the multiplicity $a$ of that eigenvalue as root of the characteristic polynomial (the geometric multiplicity here equals the algebraic multiplicity). That dimension (the rank of $\ker(T-I)$) is the complement $n-k$ of the rank of $T-I$, so one gets $a=n-k$; for the multiplicity$~b$ of $\lambda=2$ as root of the characteristic polynomial one has $a+n=b$, so $b=k$. This gives your answer.