a) Proof: If a square matrix \(A\) is diagonalizable, then there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(A=P D P^{-1}\). Hence, \[A^2 = PDP^{-1}PDP^{-1} = PD^2P^{-1}\] where \(D^2\) is a diagonal matrix as well. Hence, \(A^2\) is diagonalizable.

b) Counterexample: Consider the matrix

\[A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\]

It follows that \(A^2=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\), which is diagonalizable. However, the matrix \(A\) is not diagonalizable, as its only eigenvalue is 0 , and it is not invertible. Hence, the statement is false.