If $\dim(\operatorname{Im}T)=1$ then $T$ is diagonalizable

Question

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 !

Answer 1

$T$ 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.

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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 $x(x-\lambda)$

Answer 2

I find this to be easier to prove far more directly, but this depends on your definition of diagonalizability. As the image has dimension one, it is spanned by a single vector and thus there is $y\in V$ so that $Tx=\lambda(x)y$ for all $x\in V$. Here $\lambda(x)\in\mathbb F$ is a scalar depending on the input.

Applying the operator twice we notice that $$ T^2x = \lambda(x)\lambda(y)y $$ for all $x\in V$. Thus if $\lambda(y)=0$, we have $T^2=0$ and the operator is nilpotent. Therefore $\mu:=\lambda(y)\neq0$. Now $y$ is an eigenvector with nonzero eigenvalue $\mu$.

The map $\lambda:V\to\mathbb F$ is linear; this can be checked by using $Tx=\lambda(x)y$ and the linearity of $T$. The kernel $\ker(\lambda)$ has dimension $n-1$ — it cannot be $n$ because $y\notin\ker(\lambda)$.

Pick some vectors $b_1,\dots,b_{n-1}$ that form a basis for this kernel. Together with $y$ they form a basis $y,b_1,\dots,b_{n-1}$ of the whole $V$. Let $z,c_1,\dots,c_{n-1}\in V^*$ be the dual basis. (If $\mathbb F\in\{\mathbb R,\mathbb C\}$, you can identify $c_i=b_i$ and $z=y$ if all vectors are unit length. Then you need to add some inner products to the formulas that follow.) Because $z(y)=1$, $\lambda(y)=\mu$, and $\lambda(b_i)=z(b_i)=0$ for all $i$, we have that in fact $z=\mu^{-1}\lambda$.

Now we have found that $$ Tx = \lambda(x)y = \mu z(x)y + \sum_{i=1}^{n-1}0c_i(x)b_i. $$ This means that $T$ is diagonal in our basis with the eigenvalues $\mu,0,\dots,0$. (If this does not look like your definition of diagonalizability, you may need to adapt a little, but the same argument will work.)

Answer 3

The word "nilpotent" is missing from the title, rather confusing, since being nilpotent is easily seen to be the only possible obstruction against being diagonalisable for a rank$~1$ operator. After all, by rank-nullity this implies the nullity equals $n-1$, and this is (by definition) the dimension of the eigenspace for$~0$. Also $\operatorname{Im}(T)$ is always $T$-stable, so being of dimension$~1$ its nonzero vectors are eigenvectors, and if this is for a nonzero eigenvalue$~\lambda$, then the eigenspaces for $0$ and for $\lambda$ are complementary, and so $T$ is diagonalisable. So the only thing that can go wrong is that the nonzero vectors of $\operatorname{Im}(T)$ are eigenvectors for$~0$, but then clearly $T^2=0$ so $T$ is nilpotent.

It would have been a bit more fun if instead of $T$ not being nilpotent it had been given that $T$ has nonzero trace. Since $0$ is a root of the characteristic polynomial $\chi_T$ with multiplicity at least $n-1$ (which is its geometric multiplicity), the "final" root of $\chi_T$ equals the sum of all its roots, which is the trace of$~T$. This is also the eigenvalue of nonzero vectors in $\operatorname{Im}(T)$, and as we have seen it is zero if and only if $T$ is nilpotent if and only if $T$ is not diagonalisable.