Let $V$ be a finite dimensional vector space over a field $\mathbb{F}$ and let $T:V\to V$ be a linear operator, such that $\dim(\operatorname{Im}T)=1$ and $T$ is not nilpotent. Prove that $T$ is diagonalizable. (I denote $\dim(V)=n$).
I know that $T$ will be diagonalizable if and only if the minimal polynomial of $T$ will be a product of coprime linear factors (=algebraic multiplicity one).
From $\dim(\operatorname{Im}T)=1$ I deduce that $\exists w\in V$ such that $\forall v\in V$, $Tv=\alpha Tw $ for some scalar $\alpha \in \mathbb{F}$. In particular $Tw$ is an eigenvector (geometric multiplicity=algebraic? not sure about that). Because $\dim(\ker T)=n-1$, I deduce (not sure) that $0$ is an eigenvalue with geometric multiplicity of $n-1$. This implies the characteristic polynomial is of the form $x^{n-1}(x-\lambda)$, and because $T$ is not nilpotent, the minimal polynomial must be the same (is this correct?). I am not sure about the details here – is it true that the geometric and algebraic multiplicites are equal from the arguments I stated? Am I correct about the characteristic polynomial and the minimal polynomial?
Will be happy to receive some feedback on my arguments, and help with completing them to a formal proof.
Thank you !
Best Answer
$T$ is nilpotent if and only if its characteristic polynomial is $p_T(x) = x^n$. Here $\dim(\ker(T)) =n-1$ so you can write as you did $p_T(x)=x^{n-1}(x-\lambda)$ where $\lambda \in \mathbb{F}$, as $T$ is not nilpotent then $\lambda \neq 0$ so the algebraic multiplicity of $0$ is $n-1$, and the algebraic multiplicity of $\lambda$ is exactly one. Therefore the algebraic multiplicity of the eigenvalues is the same as their geometric multiplicity, hence $T$ is diagonalizable.
More details: The algebraic multiplicity of an eigenvalue is the number of times it appears in the characteristic polynomial, here for $0$ as $\lambda \neq0$ it appears $n-1$ times. Moreover the geometric multiplicity of $0$ is $\dim(\ker(T)) = n-1$. So the geometric and algebraic multplicity of $0$ are the same. Besides,
$$1 \leq \text{geometric multiplicity} \leq \text{algebraic multiplicity}$$
So as the algebraic multiplicity of $\lambda$ is one then its geometric multiplicity i.e. $\dim(\mathcal{E}_T(\lambda))$ is also $1$.
Finally all eigenvalues of $T$ have their geometric and algebraic multiplicity equal if and only if $T$ is diagonalizable.
And about what you said about the characteristic polynomial being the same as the minimal polynomial, it is false for $n>2$. Because when $T$ is diagonalizable, its minimal polynomial has simple roots, therefore it is $x(x-\lambda)$