Like the title says, "If an $n\times n$ matrix $A$ is diagonalizable and has only one eigenvalue $\lambda$ with multiplicity $n$, then $A = \lambda I$. True or False?"
My gut is telling me that this is true, but I'm having a little difficulty proving it formally. There's probably a very obvious proof, but thus far it has alluded me.
In any case, this is what I've done:
Since $A$ is diagonalizable, there exists a factorization such that $A = S^{-1}\Lambda S$. Since we know the eigenvalues are $\lambda$ with multiplicity $n$, $\Lambda = \text{diag}(\lambda, \lambda, …)$.
If I'm given $A = \lambda I$, it's easy to show $\lambda I = S^{-1}\Lambda S$ where $S = I$ and $\Lambda = \text{diag}(\lambda, \lambda, …)$. (Having normalized the eigenvectors). The trouble I'm having is showing $A = \lambda I$ under the given conditions.
I can simply let $S = \lambda I$, and thus $A = \lambda I$, but something about this approach feels off. (This is just my line of thought right now).
Any input would be appreciated.
Best Answer
You already have it! Since $A$ is diagonalizable, there exist an invertible matrix $S$ such that $\Lambda=\lambda I = S^{-1}AS$, so $A=S\Lambda S^{-1} = \lambda S^{-1}S = \lambda I$.