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.
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.
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).