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real symmetric matrix $B=AA^{*}$?

Let $A$ be an $n×n$ matrix with real entries. Pick out the true statements:
a. There exists a real symmetric $n × n$ matrix $B$ such that $B^2 = A^*A$.
b. If $A$ is symmetric, there exists a real symmetric $n×n$ matrix $B$ such that $B^2 = A$.
c. If $A$ is symmetric, there exists a real symmetric $n×n$ matrix $B$ such that $B^3 = A$.


I am completely stuck on this problem. How can I solve this?

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marked as duplicate by Une Femme Douce, Ittay Weiss, Micah, Did, Clive Newstead Jan 7 '13 at 7:57

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1 Answer 1

For $a$, remember that $A$ is real. What is $A^*$?

For $b$, you're trying to 'square root' the matrix. What diagonal matrix $A$ could you make so that $B$ couldn't be real?

For $c$, you can probably figure out the answer for diagonal matrices. So, with what they've given you, how can you make the matrix diagonal?

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i could not understand thehint for (c).will you explain please –  user55420 Jan 7 '13 at 6:33
    
@user55420 Real symmetric matrices are diagonalizable, so you can find a $M$ such that $M^{-1}AM$ is a diagonal matrix whose entries are the eigenvalues of $A$. Find the cube root of that matrix. Then figure out what you have to do to turn that into the cube root of $A$. –  Alexander Gruber Jan 7 '13 at 6:39

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