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Let $$A$$ be a square matrix. Proof or Counterexample.
a) If $$A$$ is diagonalizable, then so is $$A^{2}$$.
b) If $$A^{2}$$ is diagonalizable, then so is $$A$$.
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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.
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