# Minimal polynomial for a bijective linear transformation

Let $V$ be a vector space over a field $F$ which is not algebraically closed and $T:V \rightarrow V$ be a linear transformation. I am just trying to understand if injectiviity can be used to study the structre of the minimal polynomial of $T$ because none of the theorems from my linear algebra textbook make this notion explicitly clear.

So suppose first that $T$ is a bijection and that $V$ has dimension $1$. Since $T$ is a bijection can we just fix a $v \in Image (T)$ so that $Tv = av$ for some $a \in F$ and therefore the minmal polynomial of T is just $x-a$. (since $V$ has dimension 1).

Question: Suppose $dim(V) \ge 2$ if $T$ is a bijection does it follow that the minmal polynomial of $T$ has a linear factor or can we say something stronger?

It is not clear to me at this point if there is a connection between injectivity of a linear operator and diagonaliziablity .

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Are you working over an arbitrary field, or a specific one? Over $\mathbb{C}$, for example, every polynomial splits into linear factors. (And no, there is no connection between injectivity and diagonalisability.) –  Zhen Lin Aug 9 '11 at 11:29
@Zhen Thank you for the comment. I have edited to the post to rule out the trivial case of algebraically closed fields. –  user7980 Aug 9 '11 at 11:32
There's one minor connection between the two concepts: if $T$ is not injective, then $0$ is an eigenvalue of $T$ and thus the minimal polynomial has a linear factor (since $0$ always lies in the underlying field). Of course, this doesn't guarantee that $T$ is diagonalizable, but merely that it has at least one eigenvalue. –  Mark Aug 9 '11 at 11:51

If you move away from algebraically closed fields, the answer is "not in general". For example, if we work over the real field $\mathbb{R}$, the linear transformation $: \mathbb{R}^2 \to \mathbb{R}^2$ with $T(a,b) = (b,-a)$ has $T^{2} = -I$, and has minimum polynomial $x^{2}+1$, which has no linear factor in $\mathbb{R}[x].$ It is also a bijection, as is easily checked.
@Mathematica: This is possible, though computing that matrix would be the easiest way. For example, in the example above, you can see that $T^{2}(a,b) = (-a,-b)$, so $T^{2} +I =0.$ –  Geoff Robinson Jun 6 '14 at 10:27