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a) may not be true as both side are symmetric matrix but I dont know the logic.

b)true choose $B=P\sqrt{D}P^{-1}$ am I right?

c)true, chose $B=P\sqrt[3]{D}P^{-1}$ ? thank you.

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    For b) note that $\det B^2 \geq 0$, but there are symmetric $A$ for which \det A <0.2012-12-16

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Hints. (a) You should actually apply your idea for (b) (as mentioned in your question) to this part.

(b) The eigenvalues of $(PDP^{-1})^2 = PD^2P^{-1}$ must be nonnegative.

(c) Your argument sounds good, [edit] except that your $B$ defined in this way may not be real symmetric. To make $B$ symmetric, picking a $P$ that diagonalise it is not enough. You need a special $P$. (Real symmetric matrices can be diagonalised in a very special way. What is it?) The same $P$ also applies to (a) and (b).

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    @TaxiDriver $A^\ast A$ is symmetric.2013-06-17