I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method. I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra".

Question: Let $V$ be a vector space of dimension $n$ over the field $F$ and consider a linear operator $T$ on $V$ such that $\mathrm{rank}(T) = 1$. List all possible minimal polynomials for $T$.

Sketch of Proof: If $T$ is nilpotent then the minimal polynomial of $T$ is $x^k$ for some $k\leq n$. So suppose $T$ is not nilpotent, then we can argue that $T$ is diagonalizable based on the fact that $T$ must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension $1$). Thus the minimal polynomial of $T$ is just a linear term $x-c$.

Did I make a mistake assuming that $T$ can have only one nonzero eigenvalue?

Thanks for your help

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    $\begingroup$ Assume that $n\ge2$ and pick $v\ne0$ in the image of $T$, then $Tv=av$ for a given $a$ and the minimal polynomial is $x^2-ax$. $\endgroup$ – Did Jul 19 '11 at 13:14
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    $\begingroup$ If $T$ has more than one non-zero eigenvalue then it has more than one eigenvector, hence rank exceeding 1. $\endgroup$ – Gerry Myerson Jul 19 '11 at 13:18
  • $\begingroup$ Thank you Didier and Gerry so am I correct in saying that the only possible minimal polynomials are of the form $x^k$, $x-a$, and $x^2 -ax$ $\endgroup$ – user7980 Jul 19 '11 at 14:16
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    $\begingroup$ As I said: if $n\ge2$, the only possible minimal polynomial is $x^2-ax$ (for $a=0$, this includes $x^2$). $\endgroup$ – Did Jul 19 '11 at 17:02

If $n=1$ then $T$ is multiplication by some $a\in F$, the minimal polynomial $\mu_T$ of $T$ is $X-a$, and $\def\rk{\operatorname{rk}}\rk T=1$ means that $a\neq0$. Now assume $n>1$; then $T$ cannot be any scalar multiplication, so one certainly has $\deg(\mu_T)>1$.

Since $T$ has rank$~1$, it can be factored through (the $1$-dimensional vector space) $F$, that is $T=f\circ\alpha$ where $\alpha\in\mathcal L(V,F)=V^*$ and $f\in\mathcal L(F,V)$ is just scalar multiplication of a fixed vector $v\in V$. One has $T^2=f\circ\alpha\circ f\circ\alpha$, and since $\alpha\circ f\in\mathcal L(F,F)$ is scalar multiplication by $\def\tr{\operatorname{tr}}c=\alpha(v)=\tr T$, one gets $T^2=c(f\circ\alpha)=cT$, so that $\mu_T=X^2-cX=X(X-c)$. Here $c\in F$ can be anything; if $c=0$ then $T$ is nilpotent of order$~2$, hence not diagonalisable, otherwise $T$ is diagonalisable with distinct eigenvalues $0$ and $c$.

An alternative argument without using an explicit decomposition of $T$ goes as follows. By rank-nullity $T$ has an eigenvalue$~0$, so $\mu_T$ contains a factor$~X$. The remaining factor $\mu'=\mu_T/X$ of$~\mu_T$ is equal to the minimal polynomial of the restriction of$~T$ to its image. That image has dimension $\rk T=1$ on which $T$ therefore acts by some scalar$~c$, so $\mu'=X-c$, and $\mu_T=X(X-c)$. There are many ways to see that $c$ is the trace of$~T$, one of which it that the trace of any operator equals that of its restriction to its image.


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