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True or False?

Let A be a square matrix

  1. If $A$ is diagonalizable, then $A^2$ is diagonalizable.
  2. If $A$ is diagonalizable, then $A^t$ is diagonalizable.

Re 1, my answer is that it is correct, but I am a bit at a loss as to how to reason it.

Re 2, my answer is that it is correct, as $A^t=(PDP^{-1})^t=(P^{-1})^tD^tP^t=(P^t)^{-1}D^tP^t$. $D^t$ here is diagonal (in fact it is equal to $D$), so $A^t$ is diagonalizable.

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$$A\;\text{diagonalizable}\implies P^{-1}AP=D\,,\,\text{, for some invertible $P$ and diagonal}\;D\implies$$

$$D^2=(P^{-1}AP)^2=P^{-1}A^2P\implies A^2\;\;\text{diagonalizable, too}$$

Your answer to (2) is correct. You could also remember that $\;A\,,\,\,A^t\;$ are similar...

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Since $A $ is diagonalizable, $A=PDP^{-1}$, where $P$ is invertible and $D$ is diagonal matrix. So $$A^2=PDP^{-1}PDP^{-1}=PD^{2}P^{-1}.$$Since $D $ is a diagonal matrix, $D^2$ is a diagonal matrix. Hence, $A^2$ is diagonalizable.

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