Can a non-diagonal $2\times 2$ matrix with just one eigenvalue be diagonalisable?

Question

I know that some (all?) diagonal $2\times 2$ matrices can still be diagonalisable if they have just one repeated eigenvalue, but I'm wondering if it's safe to claim that a non-diagonal matrix with just one repeated eigenvalue is definitely not diagonalisable?

I'm working in $\mathrm{SL}(2,F)$ with F an algebraically closed field if that makes any difference.

Edit/Elaboration: I know about Jordan Normal Form but perhaps not enough. Let me elaborate on the bigger problem: I'm trying to show that a non-diagonal matrix in SL(2,F) with a single eigenspace is conjugate to a lower triangular matrix. Since det=1, this limits the diagonal to plus or minus 1. The paper I have read on JNF explains how to show that it's conjugate to an upper triangular matrix with the entry '1' above the diagonal. I don't think it's too hard to create an analogous argument which shows it's conjugate to a lower triangular, but adapting the entry from a '1' to any arbitrary field element seems harder.

Thanks!

Answer 1

Take for example

$$\;A=\begin{pmatrix}0&-1\\1&\;2\end{pmatrix}\implies p_A(t)=t^2-2t+1=(t-1)^2\;$$

If this matrix were diagonalizable then its minimal polynomial would be $\;t-1\;$, which clearly is false.

Thus, in general, if a matrix $\;B\;$ has a characteristic polynomial of the form $\;(t-\lambda)^2\;$ , then it is diagonalizable iff $\;t-\lambda\;$ is its minimal polynomial, which would mean that in fact

$$B=\begin{pmatrix}\lambda&0\\0&\lambda\end{pmatrix}=\lambda I$$

meaning $\;B\;$ is a scalar multiple of the identity matrix. This is in fact true, mutatis mutandis, for any square matrix with one single eigenvalue (assuming we're on an algebraically closed field)

Answer 2

If a linear transformation (at least for a finite-dimensional vector space) has only a single eigenvalue and is diagonalizable, then in some basis (and therefore all bases) it is represented by a multiple of the identity matrix.

So a non-diagonal matrix over an algebraically closed field with only a single eigenvalue cannot be diagonalizable.

Answer 3

Suppose $A$ is diagonalizable, and has only one eigenvalue $\lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = \lambda I$ such that $A = P(\lambda I)P^{-1} = \lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue. (This holds for any $n \times n$ matrix)

So we showed that diagonalizable and only one eigenvalue $\implies$ diagonal. The negation is " diagonalizable with only one eigenvalue and non-diagonal"; this cannot happen.