Corrections

[Panjete]

First of all, thanks lot for the detailed solutions and neat explanations. These have been very helpful!

Will add some potential corrections to the existing set.

Chapter 24 : Eigenvalue Problems

1. [24.1] (g) True. Since A is diagonalisable, one can write A = $X\Lambda X^{-1}$. Since all the eigen values are equal (to, say $\lambda$), $\Lambda = \lambda I$ and thus, A = $X \lambda I X^{-1} = \lambda X X^{-1} = \lambda I$, and thus A has to be diagonal. The current solution incorrectly computes the characteristic polynomial in the given counterexample.

[leeyngdo]

Thanks for your interest in my solution, and I think you're correct. I'll keep the issue open so that others can see it as well. Thanks again!