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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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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 $x^2-ax$. –  Did Jul 19 '11 at 13:14
If $T$ has more than one non-zero eigenvalue then it has more than one eigenvector, hence rank exceeding 1. –  Gerry Myerson Jul 19 '11 at 13:18
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 $x^2 -ax$ –  user7980 Jul 19 '11 at 14:16
As I said: if $n\ge2$, the only possible minimal polynomial is $x^2-ax$ (for $a=0$, this includes $x^2$). –  Did Jul 19 '11 at 17:02

1 Answer 1

If $n=1$ then $\def\rk{\operatorname{rk}}\rk T=1$ means $T\neq0$, and the minimal polynomial is $X-a$ where $T$ is multiplication by $a\neq0$. Henceforth assume $n>1$, so that by rank-nullity $T$ has an eigenvalue$~0$.

Since $X$ divides the minimal polynomial $\mu_T$, one has $\mu_T=X\mu'$ where $\mu'$ is minimal polynomial of the restriction of$~T$ to the image of $X[T]=T$. That image has dimension is $\rk T=1$, so its nonzero elements are eigenvectors of$~T$ for some eigenvalue$~a$, and $\mu'=X-a$. Hence $\mu_T=X(X-a)$ for $a\in K$ are precisely the possible minimal polynomials. The value of a is the trace of $T$, and it can be$~0$ in which case $T$ is nilpotent of order$~2$. In other cases $T$ is diagonalisable.

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