Prove/disprove that if $A$ has only one eigenvalue then it is diagonaizable only if it is a scalar matrix Given that:
$$A \in M_{n\times n} (\mathbb F), \; A \; \text{has only one eigenvalue} \; \lambda$$
We have to prove/disprove:
$$A \; \text{is diagonaizable if and only if} \; \exists \lambda : A = \lambda I$$
To be honest I've been thinking about this question for an hour now and have no clue on how to even start!
Any hints appreciated!
PLEASE take a look at the 3rd comment, that's what I got so far.
 A: You've almost got it, in the third comment above.  You just need to use that $A = PDP^{-1} = P(\lambda I)P^{-1} = \lambda PIP^{-1} = \lambda I$.
A: You only need the basic definitions for this question. An operator is diagonalisable if the (direct) sum of its eigenspaces fill the whole space (equivalently, every vector can be written as a sum of eigenvectors for distinct eigenvalues). If in addition there is only one eigenvalue$~\lambda$, then the eigenspace for$~\lambda$ must therefore fill the whole space: the operator multiplies every vector by$~\lambda$ and therefore equals$~\lambda I$ (or: every vector is the sum of at most one eigenvector for$~\lambda$, so it either is such an eigenvector or it is$~0$).
Of course the opposite direction that $\lambda I$ is diagonalisable is trivial.
A: An easier approach...but knowing some slightly more advanced stuff:
== $\,A\,$ has one single eigenvalue $\,\lambda\,$ iff its characteristic polynomial is $\,(x-\lambda)^n\,$
== A matrix is diagonalizable iff its minimal polynomial is a product of different linear factors, from which it follows that
$$A\;\text{is diagonalizable}\iff\;(x-\lambda)\;\text{is its minimal polynomial}\iff A-\lambda I=0\iff A=\lambda I$$
