Does diagonalizable imply distinct eigenvalues?
Solution
No. For \(n\ge2\), the identity matrix \[ I_n \] is already diagonal, hence diagonalizable, but its characteristic polynomial is \[ \chi_{I_n}(x)=(x-1)^n. \] Thus it has only the single eigenvalue \(1\), with algebraic multiplicity \(n\).
An \(n\times n\) matrix with \(n\) distinct eigenvalues is diagonalizable, since eigenvectors for distinct eigenvalues are linearly independent; \(I_n\) shows that the converse fails.