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Let $A$ be an $n\times n$ positive definite matrix. Show that there exists a unique positive definite matrix $B$ such that $B^2=A$.

I do know the existence. But what about the uniqueness? Would you help me out? Thank you.

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Suppose $B_1=UDU^\ast$ and $B_2=V\Lambda V^\ast$ are two positive definite square roots of $A$, where $U$ and $V$ are unitary and $D,\Lambda$ are positive diagonal matrices. Since both $D$ and $\Lambda$ contain the positive square roots of the eigenvalues of $A$, the two matrices must be permutation similar. Therefore, by absorbing some appropriate permutation matrices into $U$ and $V$, we may assume WLOG that $D=\Lambda=(\lambda_1 I_{k_1})\oplus\cdots\oplus(\lambda_r I_{k_r})$, where $\lambda_1,\ldots,\lambda_r$ are distinct. Now the equality $B_1^2=B_2^2$ implies that $D$ commutes with $W=V^\ast U$. Hence $W$ must be a block diagonal matrix whose partitioning conforms to the block structure of $D$. But then $WDW^\ast=D$ and hence $B_1=UDU^\ast=VWDW^\ast V^\ast=VDV^\ast=B_2$.

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Being symmetric, $A$ is diagonalisable. Any square root $B$ commutes with $A$, so it must stabilise the eigenspaces of $A$. The restriction of $B$ to the eigenspace $V$ of $A$ for $\lambda>0$ is a symmetric positive definite square root of the restriction $\lambda I_V$ of $A$, so it suffices to show that $\sqrt\lambda I_V$ is the unique such square root.

But the restriction of $B$ is diagonalisable, any eigenvalue it has must be a square root of$~\lambda$ and also positive; a diagonalisable matrix with $\sqrt\lambda$ as unique eigenvalue cannot be other than $\sqrt\lambda I_V$.

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Hints:

Can you write the matrix in Jordan Normal Form (why)?

If you have $B = P D^{1/2}P^{-1}$, where $D$ (of course, this can be complex valued) represents the square root of the diagonal matrix (which is just the square root of the eigenvalues), what does $B^2$ equal?

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  • 1
    $\begingroup$ Could you please elaborate a little bit? $\endgroup$ – Leo Mar 18 '15 at 17:01

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