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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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    What is $\sqrt{D}$ when the entries are negative?2012-12-16
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    ah! I see, thank u2012-12-16
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    You're not done yet. What about a)?2012-12-16
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    $detA=detB$ we need? and $b$, $c$ are false?2012-12-16
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    Your title and text body don't match.2012-12-16
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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

1 Answers 1

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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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    could you please tell me how to solve $a$ precisely? I am not done yet, $\det B^2=\det A^{*}\times \det A$ is that always happen?2013-06-17
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    @TaxiDriver $A^\ast A$ is symmetric.2013-06-17