031 Review Part 1, Problem 2

True or false: If a matrix  $A^{2}$  is diagonalizable, then the matrix  $A$  must be diagonalizable as well.

Solution:
Let  $A={\begin{bmatrix}0&1\\0&0\end{bmatrix}}.$
First, notice that
$A^{2}={\begin{bmatrix}0&0\\0&0\end{bmatrix}},$
which is diagonalizable.
Since  $A$  is a triangular matrix, the eigenvalues of  $A$  are the entries on the diagonal.
Therefore, the only eigenvalue of  $A$  is  $0.$  Additionally, there is only one linearly independent eigenvector.
Hence,  $A$  is not diagonalizable and the statement is false.

Final Answer:
FALSE

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